DAILY ASSIGNMENTS
Week 1 (First day for U of I classes is Tuesday, January 17, 2012)
Jan 17 (Tue) Go to your discussion-recitation section.
Jan 18 (Wed) Read sections 1.1 and 1.2. In 1.1 do #4, 7, 8, 25, 34, 41, 47, 51, 54, 71, 73, 77. In 1.2 do #5, 10, 16, 18. You should complete these before your next discussion-recitation section meeting. Be sure that you are prepared for this course – if you have not received a 70% or higher on the ALEKS math placement test some time between September 15, 2011 and January 23, 2012, then you will be automatically dropped from the course.
Jan 20 (Fri) Read section 1.3. In 1.3 do #3, 8, 9, 10, 12, 14, 18, 19, 31, 32, 33, 38, 41, 43. We determined the domain for the following functions.
• f(x) = x2 + 3
• y = 1 / (2x − 5)
• g(t) = sqrt(8 − 2t)
• h(x) = sqrt(x2 − 4x + 3)
Using the definition of even and odd functions, we determined whether the following functions are even, odd or neither.
• f(x) = x2 is an even function.
• g(x) = x3 is an odd function.
• h(t) = 3t5 + 5 is neither even nor odd.
• w(x) = (3x2 + 1)3 is an even function.
• v(t) = 3cos(t) + t4 is an even function.
• p(x) = 2sin(x) + x5 is an odd function.
For the first three functions above we also used the graphical approach and saw the symmetry in even functions versus odd functions.

Students should know how to graph basic functions such as the following.

• x, x2, x3, x4, ...
• 1/x, 1/x2, 1/x3, 1/x4, ...
• x1/2, x1/3, x1/4, ...
• ex, ln(x)
• sin(x), cos(x), tan(x)
I showed how to graph all of these except the last line of trigonometric functions, but you will need to quickly graph any of these functions without a calculator. I'll discuss trigonometric functions in great detail on Monday.

I used shifting techniques to graph y = 3(x − 2)2 + 5 by slowly modifying the graph of y = x2. Students should learn to shift any of the basic functions to obtain graphs of more complicated functions.

Week 2
Jan 23 (Mon) For homework finish Trigonometry Worksheet #1 and read Appendix D.

In lecture I discussed some basic trigonometry. After today's lecture you should

• Know the definition of cos(θ) and sin(θ) as the x and y coordinates of points along the unit circle.
• Be able to easily switch between degrees and radians.
• Be able to evaluate cos(θ) and sin(θ) for special angles.
• Be able to easily recall these or similar trigonometric identities by thinking about the definition of cos(θ) and sin(θ) in terms of the unit circle.
• cos(−θ) = cos(θ)   (thus cosine is an even function)
• sin(−θ) = −sin(θ)   (thus sine is an odd function)
• cos(θ + 2π) = cos(θ)
• sin(θ + 2π) = sin(θ)
• cos(π − θ) = −cos(θ)
• sin(π − θ) = sin(θ)
• cos(π/2 − θ) = sin(θ)
• sin(π/2 − θ) = cos(θ)
• sin2(θ) + cos2(θ) = 1
• Be able to graph y = sin(x) and y = cos(x). Know the x-intercepts along with the domain and range for both functions.
• Know the other trigonometric functions in terms of sine and cosine. That is,
• tan(θ) = sin(θ) / cos(θ)
• cot(θ) = cos(θ) / sin(θ)
• sec(θ) = 1 / cos(θ)
• csc(θ) = 1 / sin(θ)
• Be able to evaluate tan(θ), cot(θ), sec(θ) and csc(θ) for special angles by first writing everything in terms of sine and cosine.
• Be able to graph y = tan(x), y = cot(x), y = csc(x) and y = sec(x). At first it will help to rewrite everything in terms of sine and cosine.
• Know the identities tan2(θ) + 1 = sec2(θ) and cot2(θ) + 1 = csc2(θ). These are easily derived by dividing both sides of the equation sin2(θ) + cos2(θ) = 1 by either cos2(θ) or sin2(θ).
See the schedule of free tutoring which begins tonight. Note that my own office hours have been modified for this week only.
Jan 25 (Wed) Skip section 1.4 and read section 1.5. For homework do #29, 30, 35, 36, 37, 38, 65, 67 in Appendix D. You will need a calculator for #35–38. Then do #2, 4, 11, 12, 13, 14, 15, 16, 17, 19, 20, 21, 22, 25, 29, 30 in section 1.5. There will be two quizzes next week – quiz #1 on Tuesday and quiz #2 on Thursday. Quiz #1 will cover sections 1.1, 1.2, 1.3 and problems on trigonometry (see worksheet and Appendix D). Quiz #2 will cover sections 1.5 and 1.6.

I discussed composition of functions. The domain of (f o g)(x) is all x in the domain of g for which g(x) is in the domain of f. The key thing to remember here is that (f o g)(x) = f(g(x)) and we evaluate the inside function g(x) first. For example if f(x) = x2 + 3 and g(x) = sqrt(x − 2) then (f o g)(x) = f(g(x)) = f(sqrt(x − 2)) = (sqrt(x − 2))2 + 3 = x − 2 + 3 = x + 1. Even though we can plug any x-value into the expression x + 1, the domain of (f o g)(x) is not all real numbers. The domain of (f o g)(x) is [2, ∞) since we need to evaluate g(x) first.

I showed how to graph y = tan(x) and y = sec(x). You should use this approach or a similar approach to graph y = csc(x) or y = cot(x).

In a right triangle where θ is one of the acute angles we label the length of the side opposite θ as opp, the length of the side adjacent θ as adj, and the length of the hypotenuse as hyp. In lecture I used similar triangles and the definition of cosine and sine on the unit circle to obtain the following relationships in a right triangle.

• sin(θ) = opp / hyp
• cos(θ) = adj / hyp
• tan(θ) = opp / adj
• csc(θ) = 1 / sin(θ) = hyp / opp
• sec(θ) = 1 / cos(θ) = hyp / adj
• cot(θ) = 1 / tan(θ) = adj / opp
I used right triangle trigonometry to solve problems like the following:
• Given that 0 < θ < π/2 and cos(θ) = 3/4, determine the value of cot(θ).
• Given that 0 < θ < π/2 and sin(θ) = x/2, determine the value of tan(θ).
• If one of the three interior angles in a right triangle is π/5 radians and the side opposite this angle has length 2, then what is the length of the hypotenuse?
I then talked about recognizing linear versus exponential functions from a table of values.

If the x-value entries in a table of values are incremented by a constant amount, then the following holds.

• For linear functions, to go from one y-value to the next you add a constant amount.
• For exponential functions, to go from one y-value to the next you multiply by a constant amount.
We determined that an exponential function y = C*ax (a > 0, a ≠ 1) fits the first set of data, a linear function y = mx + b fits the second set of data, and no linear or exponential function exactly fits the third set of data. For homework find formulas for functions which fit the first two sets of data.
• TABLE 1
x y
0 3
2 12
4 48
6 192
• TABLE 2
x y
0 3
3 18
6 33
9 48
• TABLE 3
x y
0 2
5 8
10 64
15 1024
For homework find a formula for a function which fits the following set of data. Determining whether it could be linear or exonential is the easy part. Finding the formula is more difficult than the examples done in class since we do not begin with x = 0.
• TABLE 4
x y
1 2
3 4
5 8
7 16
Jan 27 (Fri) Read section 1.6. In 1.6 do #5, 7, 9, 10, 15, 17, 19, 21–26, 35–41, 51–54, 57, 58. Tuesday's quiz will be on sections 1.1, 1.2, 1.3 and problems on trigonometry (see worksheet and Appendix D which both now have solutions). Thursday's quiz will cover sections 1.5 and 1.6.

Students should have obtained the following formulas for functions which fit the data given in TABLE 1, TABLE 2 and TABLE 4 from last lecture.

TABLE 1: By plugging (x,y) = (0,3) and (x,y) = (2,12) into y = C*ax we obtained the formula y = 3*2x.

TABLE 2: By plugging (x,y) = (0,3) and (x,y) = (3,18) into y = mx + b we obtained the formula y = 5x + 3.

TABLE 4: By plugging (x,y) = (1,2) and (x,y) = (3,4) into y = C*ax we obtained the two equations 2 = C*a and 4 = C*a3. We saw two different ways to solve these equations and obtain a = sqrt(2) and C = sqrt(2) resulting in the formula y = sqrt(2)*sqrt(2)x. There are other ways to write this formula.

For y = C*ax we have exponential growth if a > 1 and exponential decay if 0 < a < 1. We looked at the graphs of examples y = 10*2x and y = 100*(1/2)x.

We looked at a population which is 50 in the year 1980 and doubles every 10 years after that. When does the population reach 600? A quick table shows that this occurs somewhere between 30 and 40 years after 1980.

# years after 1980 population
t P
0 50
10 100
20 200
30 400
40 800

To determine a more precise answer we plugged (t,P) = (0,50) and (t,P)=(10,100) into P = C*at to obtain the formula P = 50*2t/10. Setting P = 600 and solving for t requires logarithms. We get t = 10*ln(12) / ln(2) ≈ 35.8 years after 1980. Students should become comfortable using the rules of logarithms.

I introduced inverse functions by first looking at the following tables of values for y = x3 and y = x1/3.

x y = x3
−2 −8
−1 −1
0 0
1 1
2 8
3 27
4 64

x y = x1/3
−8 −2
−1 −1
0 0
1 1
8 2
27 3
64 4

The roles of x and y are reversed for inverse functions such as f(x) = x3 and f −1(x) = x1/3. Be careful with this misleading but standard notation since f −1(x) is not the same as 1 / f(x).

We saw that the graph of f(x) and f −1(x) are mirror images of each other across the line y = x.

Thinking about the relationship between a function and its inverse, one can quickly see that g(x) = x + 8 has inverse g −1(x) = x − 8, and that h(x) = 10x has inverse h −1(x) = x/10. For more complicated functions such as f(x) = (4x − 3)1/5 it isn't as immediately apparent that f −1(x) = (x5 + 3) / 4. We looked at two ways of obtaining the formula for the inverse of this function.

We discussed the concept of one-to-one functions and the horizontal line test as a graphical way to see whether or not a function is one-to-one and thus has an inverse. We saw that for functions like f(x) = x2 which are not one-to-one, we could restrict the domain so that it has an inverse.

I went over the basic definition of the logarithm for any base along with the various rules for manipulating logarithms. In particular

• loga(x) = y ⇔ ay = x (note that ⇔ means if and only if)
• loga(ax) = x
• aloga(x) = x for x > 0
• loga(x*y) = loga(x) + loga(y)
• loga(x/y) = loga(x) − loga(y)
• loga(xr) = r*loga(x)
Week 3 (Deadline to add this course is Monday, January 30, 2012)
Jan 30 (Mon) For homework read sections 2.1 and 2.2. In 2.1 do #5. In 2.2 do #4, 7, 8, 11, 15, 17, 23, 24, 29, 31, 32, 39, 41. Quiz #1 on Tuesday will cover sections 1.1, 1.2, 1.3 and problems on trigonometry (see worksheet and Appendix D). Quiz #2 on Thursday will cover sections 1.5 and 1.6.

A function is called one-to-one if it never takes on the same value twice; that is, f(x1) ≠ f(x2) whenever x1 ≠ x2. The horizontal line test is a geometric way to understand the term one-to-one. All one-to-one functions have inverses. Since increasing functions are one-to-one, an increasing function always has an inverse. Since decreasing functions are one-to-one, a decreasing function always has an inverse.

The function f(x) = x3 + x5 is increasing so it has an inverse. Our technique to find a formula for the inverse did not work since we were unable to solve x = y3 + y5 for y. However, the function still has an inverse and we know things about it. For instance, since (1, 2) is on the graph of y = f(x), we know that the point (2, 1) is on the graph of y = f −1(x).

We spent a little time on logarithms with various bases. We spent more time on natural logarithms which have base e ≈ 2.71828...

The function f(x) = ex has inverse f −1(x) = ln(x). You should know the following definitions, identities and simplification rules.

• ln(x) = y ⇔ ey = x (note that ⇔ means if and only if)
• ln(ex) = x for all x
• eln(x) = x for all x > 0
• ln(x*y) = ln(x) + ln(y)
• ln(x/y) = ln(x) − ln(y)
• ln(xr) = r ln(x)
• ln(1) = 0
• ln(e) = 1
• e0 = 1
From the definition of ln(x) as the inverse function to ex so that ln(x) = y ⇔ ey = x, we quickly see the following.
• ln(e3) = 3
• ln(1/e) = −1 since e−1 = 1/e
• ln(sqrt(e)) = 1/2 since e1/2 = sqrt(e)
• ln(1) = 0 since e0 = 1
• ln(e) = 1 since e1 = e
This technique can also be used to prove identities.
• ln(x*y) = ln(x) + ln(y) since eln(x) + ln(y) = eln(x)*eln(y) = x*y
• ln(x/y) = ln(x) − ln(y) since eln(x) − ln(y) = eln(x) / eln(y) = x/y
• ln(xr) = r ln(x) since er ln(x) = ( eln(x) )r = xr
We solved for x in the equation 3 = 24x − 1 by taking the natural logarithm of both sides and using the appropriate identity.

We solved for x in the equation 6 + 2x = 4 + 2x + 2. Taking the logarithm of both sides didn't work directly since we do not have a rule for taking the logarithm of a sum. We had to manipulate the equation appropriately before taking the logarithm of both sides.

We used the informal definition of limit as found on the first page of section 2.2 to determine the value of the following limits.

1. limx → 3 ( x2 ) = 9
2. limx → 4 ( (x2 − 2x − 8) / (x − 4) ) = 6
3. limx → 0 ( sin(x) / x ) = 1
4. limx → π/2 ( tan(x) ) (we need to break it up into two one-sided limits)
• limx → π/2 ( tan(x) ) = ∞
• limx → π/2 + ( tan(x) ) = −∞
We determined these limits in many ways.
1. First we graphed y = x2 to see that the y-values are approaching 9 as x approaches 3. Next we made a table of values for y = x2 using x-values 2.9, 2.99 and 2.999 as well as 3.1, 3.01 and 3.001. We saw in our table that the y-values seem to be approaching 9 as x approaches 3. Of course most students expect y = x2 to approach 32 or 9 as x approaches 3.
2. First we made a table of values for y = (x2 − 2x − 8) / (x − 4) using x-values 3.9, 3.99 and 3.999 as well as 4.1, 4.01 and 4.001. We saw in our table that the y-values seem to be approaching 6 as x approaches 4. Next we factored the numerator and rewrote y = (x2 − 2x − 8) / (x − 4) = (x − 4)(x + 2) / (x − 4) = x + 2. In this form we would expect y = x + 2 to approach 6 as x approaches 4. Lastly we looked at the graph of y = (x2 − 2x − 8) / (x − 4) which looks just like the graph of y = x + 2 with a hole in the graph at the point (4, 6). From this graph, we see that the y-values are approaching 6 as x approaches 4.
3. First we made a table of values for y = sin(x) / x using x-values −0.1, −0.01 and −0.001 as well as 0.1, 0.01 and 0.001. As long as we used radian measure, we saw in our table that the y-values seem to be approaching 1 as x approaches 0. I will prove this soon.
4. For this we relied solely on the graph of y = tan(x) to see that the limit from the left is ∞ while the limit from the right is −∞.
Jan 31 (Tue) Quiz #1 on sections 1.1, 1.2, 1.3 and trigonometry will be given during today's discussion section.
Feb 1 (Wed) Everyone should read section 2.3. Math majors or those who want a better understanding of proof techniques for limits should read section 2.4. In section 2.3 do #11, 13, 15, 17, 18, 20, 25, 26, 37, 39. There will be a quiz tomorrow on sections 1.5 and 1.6.

I went over #7 from the homework in section 2.2. We used that limx → a f(x) = L if and only if limx → a f(x) = L and limx → a+ f(x) = L.

We used the informal definition of limit as found on the first page of section 2.2 to determine the value of the following limits.

1. limx → π/2 ( tan(x) ) (we need to break it up into two one-sided limits)
• limx → π/2 ( tan(x) ) = ∞
• limx → π/2 + ( tan(x) ) = −∞
2. limx → 7 ( (sqrt(x + 2) − 3)/(x − 7) ) = 1/6
We determined these limits in different ways.
1. For this we relied solely on the graph of y = tan(x) to see that the limit from the left is ∞ while the limit from the right is −∞.
2. For this we multiplied the numerator and denominator each by sqrt(x + 2) + 3 and simplified. This made it easier to guess that the limit would be 1/6.
We saw numerical evidence last time to suggest that limθ → 0 ( sin(θ) / θ ) = 1. I proved this by using the unit circle and comparing the areas of a small triangle, the sector of a circle and a large triangle.

We next looked at an example where S(t) = 2t gives the size of a tumor in cubic millimeters t months after its discovery. In order to determine how quickly the size of the tumor is increasing precisely 6 months after its discovery, we ended up taking the following limit.

• S ′(6) = limt → 6 ( (2t − 26) / (t − 6) )

This can also be written as

• S ′(6) = limh → 0 ( (26 + h − 26) / (h) )

We made a table of values to approximate this limit. I mentioned that this S ′(6) notation for the rate at which the tumor is growing at precisely t = 6 is called the derivative. Since some students have learned a little about derivatives before, I asked if anyone knew a formula for S ′(t) so that we could simply plug t=6 into this formula. Most of the students that have seen derivatives before obtained an incorrect formula for S ′(t). I used this example to point out the need for understanding the method I am using with tables of values and other techniques before trying to use short-cut techniques.

We next evaluated the following limits.

1. limx → 0 ( 3 + 2e−x ) = 5
2. limx → ∞ ( 3 + 2e−x ) = 3
3. limx → −∞ ( 3 + 2e−x ) = ∞
4. limx → 0 x2 sin(1 / x) = 0
I graphed y = 3 + 2e−x to obtain the first three limits above. For the last limit, we saw that −1 ≤ sin(1/x) ≤ 1 so that −x2 ≤ x2sin(1/x) ≤ x2. This limit was then found using The Squeeze Theorem.
Feb 2 (Thu) Quiz #2 on section 1.5 and 1.6 will be given during today's discussion section.
Feb 3 (Fri) Read section 2.5 and reread the material from section 1.6 on inverse trigonometric functions. In section 1.6 do #63, 64, 65, 68, 70, 71. In section 2.5 do #20, 45, 49, 51, 53.

You should know the following identities.

• sin(2θ) = 2sin(θ)cos(θ)
• cos(2θ) = cos2(θ) − sin2(θ) which can also be written in the following two ways using the identity sin2(θ) + cos2(θ) = 1.
• cos(2θ) = 2cos2(θ) − 1
• cos(2θ) = 1 − 2sin2(θ)
We evaluated the following limits.
1. limθ → 0 sin(2θ) / (5θcos(θ)) = 2/5 (see reasons below)
2. limx → 3+ ln(x − 3) = −∞
3. limx → 4 (x + 2) / (x − 4) = −∞
4. limw → 1 (w4 − 1)/(w − 1) = 4
For limit (1) we used the identity sin(2θ) = 2sin(θ)cos(θ), cancelled the cos(θ), and then used limθ → 0 sin(θ) / θ = 1. For limit (2) we graphed y = ln(x − 3) to quickly obtain an answer. For limit (3) we rewrote (x + 2) / (x − 4) = (x − 4 + 6) / (x − 4) = (x − 4) / (x − 4) + 6 / (x − 4) = 1 + 6 / (x − 4) and then graphed the function to obtain an answer. I discussed how we should have expected the answer for limit (3) immediately without graphing or rewriting the expression. For limit (4) we factored the numerator and cancelled terms in order to obtain an answer.

Although it is good to use common sense, we saw from a few examples that what we think of as common sense may not actually be correct in a given situation. For instance, even though limx → 0 ( x2 ) = 0, we do not immediately know the value of limx → 0 ( x2 * f(x)). We looked at f(x) = 5x + 2, f(x) = 1/x6, and f(x) = (6x + 8) / x2 to see that limx → 0 ( x2 * f(x)) might equal 0, ∞, 8, or some other value depending upon the choice for f(x).

We define f to be continuous at x = a if limx → a f(x) = f(a).

Nearly every function we will use in calculus is continuous on its domain so we can often just plug in a particular x-value to determine a limit. In particular, polynomials, exponentials, logarithms, roots, trig functions, inverse trig functions and rational functions are all continuous on their domains. We can also combine continuous functions by adding, subtracting, multiplying or dividing and obtain another continuous function. We can even do composition of functions to get another continuous function. For dividing, just make sure that the denominator is not 0. For composition of functions see Theorem 9 from section 2.5.

We found limx → 2 (3x2 + 5x − 4) = 18. Using the limit laws from section 2.3 it takes multiple steps to find this limit. However, knowing that polynomials are continuous everywhere it is much easier to simply plug in x = 2 to obtain the limit in one step.

We then discussed the very important Intermediate Value Theorem. We used it to explore the location of the roots (i.e. x-intercepts) for f(x) = x3 − 3x2 − x + 5. For quizzes and tests, I expect you to be able to carefully write out the statement of important theorems such as this one or The Squeeze Theorem. You should also be able to carefully show how and why you may apply these theorems to specific problems.

In lecture we looked carefully at why we restrict the domain of f(x) = sin(x) in order for it to have an inverse function f −1(x) = sin −1(x) = arcsin(x). Note that sin −1(x) is not the same as 1 / sin(x). Students should also understand how the domains of cos(x), tan(x), and sec(x) are restricted in order to have inverse functions. We then evaluated the first of the following three quantities. For next time think carefully about the values of the last two quantities.

• sin −1(−1/2)
• sin(sin −1(1/5))
• sin −1(sin(3π/4)) (be very careful with this one — the answer is not 3π/4)
Week 4
Feb 6 (Mon) Read section 2.6. In section 2.6 do #8, 15, 21, 24, 25, 29, 30, 33, 41, 43. There will be a quiz Thursday on sections 1.6, 2.1, 2.2, 2.3, 2.5 and 2.6. The test one week from Wednesday will cover sections 1.1, 1.2, 1.3, 1.5, 1.6, 2.1, 2.2, 2.3, 2.5, 2.6, 2.7, 2.8 and the trigonometry material discussed in lecture.

The logarithm buttons found on most calculators are ln (natural logarithm with base e) or log (common logarithm with base 10), so for other logarithms it is useful to convert to one of these bases. We note that y = log2 13 ⇔ 2y = 13 so we expect y to be between 3 and 4. Taking the logarithm of both sides of this second equation and solving for y gives y = ln(13) / ln(2) ≈ 3.7. We can use this same technique to prove the change of base formula shown below.

• loga x = ln(x) / ln(a)
Does common sense help us determine limx → ∞ ( 1 + 1/(2x) )x ? Many students expect the limit to be either 1 or ∞ depending upon whether they concentrate on the term in parentheses or the exponent. We will see later that surprisingly this limit is equal to the square root of e.

We then discussed an approach for finding limx → ∞ f(x) or limx → −∞ f(x) where f(x) is the quotient of either polynomials or roots of polynomials. We can rewrite the quotient by dividing both the numerator and the denominator by xn where n is the highest power of x found in the denominator. We looked at the following examples. Think carefully about the last one.

• limx → ∞ (2x2 + 1) / (5 + 3x2) = limx → ∞ (2 + 1/x2) / (5/x2 + 3) = 2/3
• limx → −∞ (9 + x + 4x2) / (x5 + 6) = limx → −∞ (9/x5 + 1/x4 + 4/x3) / (1 + 6/x5) = 0
• limx → ∞ 8x10 / (2x3 + 11) = limx → ∞ 8x7 / (2 + 11/x3) = ∞
• limx → ∞ (15x3 + 1) / sqrt(4x6 + 2x + 5) = limx → ∞ (15 + 1/x3) / sqrt(4 + 2/x5 + 5/x6) = 15/2
Although you may know short-cuts for obtaining the previous limits, I would like you to use an approach like I've used above to demonstrate why each limit is equal to the given values.

Last time we defined f to be continuous at x = a if limx → a f(x) = f(a).

We looked at the piecewise function defined so that f(x) = x2 + C for x ≤ 2 and f(x) = 3 − x for x > 2. Which value for C makes f(x) continuous everywhere? Since continuity is defined in terms of limits, you should use limits in your solution.

It is important to know how to restrict the domains of the trigonometric functions so that they are one-to-one and thus have inverse functions. I showed the graph of sin(x) for −π/2 ≤ x ≤ π/2 along with the graph of sin −1(x). I discussed how the domain and range are related for the two functions. Students should work on their own to understand how to restrict the domains of cos(x), tan(x) and sec(x) to obtain cos −1(x), tan −1(x) and sec −1(x). We then evaluated the following quantities.

• sin(sin −1(1/5))
• sin −1(sin(3π/4)) (be very careful with this one — the answer is not 3π/4)
• tan −1(sqrt(3))
• cos(sin −1(1/3))
For this last example we let θ = sin −1(1/3) so that sin(θ) = 1/3 and −π/2 ≤ θ ≤ π/2. We then drew a right triangle for which the side opposite θ equals 1 and the hypotenuse equals 3. Pythagorean's theorem then quickly allowed us to obtain cos(sin −1(1/3)) = cos(θ) = sqrt(8)/3. We'll see a more formal way using trigonometric identities to arrive at this answer next time.
Feb 8 (Wed) Read sections 2.7 and 2.8. In section 2.7 do #5, 6, 7, 8, 9, 10, 13, 14, 27, 28, 29, 30, 31, 32. In section 2.8 do #4, 5, 6, 12, 16, 17, 18, 21, 23, 25, 27, 29. Prepare for tomorrow's quiz on sections 1.6, 2.1, 2.2, 2.3, 2.5 and 2.6. The test one week from today will cover sections 1.1, 1.2, 1.3, 1.5, 1.6, 2.1, 2.2, 2.3, 2.5, 2.6, 2.7, 2.8 and the trigonometry material discussed in lecture.

I discussed the restricted domains for tan(x) and cos(x) so that they are one-to-one and have inverses cos −1(x) and tan −1(x), respectively. We also looked at the graphs of these inverse functions. We then evaluated the following quantities.

• cos(sin −1(2/3))
• cot(cos −1(x))
• sin(2*tan−1(1/5))
I showed how to use the cos2 θ + sin2 θ = 1 identity along with knowledge of restricted domains and the unit circle to evaluate the first two quantities above. I also showed the faster approach of drawing a particular right triangle. We found sin(2*tan−1(1/5)) by letting θ = tan−1(1/5) or tan(θ) = 1/5 and drawing a right triangle with opp / adj = 1 / 5. With Pythagorean's Theorem we found the hypotenuse to be sqrt(26). This resulted in

sin(2*tan−1(1/5)) = sin(2*θ) = 2*sin(θ)*cos(θ) = 2*(1/sqrt(26))*(5/sqrt(26)) = 10/26 = 5/13.

I discussed how limits are used to determine horizontal and vertical asymptotes. We looked at graphs which showed why using limits is appropriate. We found limx → 2+ ( 3 / (x − 2) ) = ∞ so that the graph of f(x) = 3 / (x − 2) has a vertical asymptote at x = 2.

I showed how to choose which technique to use when finding the following limits.

• limx → 2 (3x2 − 5x − 2) / (2x2 − 5x + 2) = 7/3
• limx → 0.5+ (3x2 − 5x − 2) / (2x2 − 5x + 2) = ∞
• limx → ∞ (3x2 − 5x − 2) / (2x2 − 5x + 2) = 3/2
I defined the derivative of f(x) as

f ′(x) = limh → 0 (f(x + h) − f(x)) / h

We saw how this definition gives us the slope of the tangent line at a point.

We used the limit definition of a derivative to show that f(x) = x2 has derivative f ′(x) = 2x.

Feb 9 (Thu) Quiz #3 on sections 1.6, 2.1, 2.2, 2.3, 2.5 and 2.6 will be given during today's discussion section.
Feb 10 (Fri) Prepare for Wednesday's test on sections 1.1, 1.2, 1.3, 1.5, 1.6, 2.1, 2.2, 2.3, 2.5, 2.6, 2.7, 2.8, and the trigonometry material discussed in lecture. No calculators or notes are allowed, and you should bring a student ID. The test will be given during your officially scheduled lecture period. See the notes written here after each lecture to get a summary of some of the important material. It is expected that you now know how to solve every assigned homework problem. Detailed solutions to all odd problems and assigned even problems are available at Illinois Compass 2g. Solutions to the quizzes and the trigonometry worksheet are posted on the course homepage. You must be able to state and use the definitions of even functions, odd functions, continuity, derivatives. You must be able to state and use the Intermediate Value Theorem, Squeeze Theorem, and the theorem which says If f is differentiable at a, then f is continuous at a. You will definitely have one problem where you will be asked to find the derivative of a function using limits. Use proper notation as you show all the appropriate steps.

You may want to look over old tests and quizzes from my previous MATH 220 courses. Check the available tutoring hours. No appointment is necessary.

The cover page on your test will include a seating chart. When you pick up your test on Wednesday, one of these seat numbers will be circled and that will be your assigned seat for the first test. Try to arrive early to obtain your assigned seat. Unless you happen to sit at the end of a row, you will be required to stay for the full 50 minute testing period. The TAs will not answer any questions during the test.

We used the graph of a function y = f(x) to determine a reasonable graph for the derivative function y = f ′(x). We saw that graph of the derivative of sin(x) looks like cos(x) and the graph of the derivative of ex looks like ex. Later we will prove that these are actually the correct formulas for these derivatives.

We saw graphically why the following three definitions of derivative are equivalent.

1. f ′(x) = limh → 0 (f(x + h) − f(x)) / h
2. f ′(x) = limw → x (f(w) − f(x)) / (w − x)
3. f ′(a) = limx → a (f(x) − f(a)) / (x − a)
We used the 2nd approach to find that the derivative of f(x) = x3 is f ′(x) = 3x2. For homework you should use the 1st approach to see that you obtain the same formula for the derivative of f(x) = x3. Usually one just uses whichever approach seems easiest for a particular function.

I mentioned that the derivative, the rate of change, and the slope all represent the same quantity.

We looked at a population P(t) = 2000 + 3t2. We calculated P(5) = 2075 and P ′(5) = 30 to show that 5 years later the population is 2075 people and increasing by 30 people per year.

We looked at the height of a ball thrown upwards from an apartment window h(t) = −16t2 + 96t + 160. We saw that h(0) = 160 feet is the height of the window. We graphed h(t) to see it had a slope of 0 when the ball reached its maximum height. One can use limits to find the velocity h ′(t) = −32t + 96. We set h ′(t) = 0 to determine that the ball reached its maximum height at t = 3 seconds. With this information we can determine h(3) = 304 feet to be the maximum height. We set h(t) = 0 to find how long it takes until the ball falls back to the ground. We plugged this value of t into the velocity formula to obtain the velocity at the moment the ball hit the ground. I discussed why we expected the velocity to be negative.

I mentioned the important theorem which states that if a function is differentiable at a point, then it must also be continuous at that point. I'll prove this quickly on Monday.

For most of Monday's lecture I will answer student questions in preparation for Wednesday's test.

Week 5
Feb 13 (Mon) Prepare for Wednesday's test on sections 1.1, 1.2, 1.3, 1.5, 1.6, 2.1, 2.2, 2.3, 2.5, 2.6, 2.7, 2.8, and the trigonometry material discussed in lecture. No calculators or notes are allowed, and you should bring a student ID. The test will be given during your officially scheduled lecture period. See the notes written here after each lecture to get a summary of some of the important material. It is expected that you now know how to solve every assigned homework problem. Detailed solutions to all odd problems and assigned even problems are available at Illinois Compass 2g. Solutions to the quizzes and the trigonometry worksheet are posted on the course homepage. You must be able to state and use the definitions of even functions, odd functions, continuity, derivatives. You must be able to state and use the Intermediate Value Theorem, Squeeze Theorem, and the theorem which says If f is differentiable at a, then f is continuous at a. You will definitely have one problem where you will be asked to find the derivative of a function using limits. Use proper notation as you show all the appropriate steps.

You may want to look over old tests and quizzes from my previous MATH 220 courses. Check the available tutoring hours. No appointment is necessary.

The cover page on your test will include a seating chart. When you pick up your test on Wednesday, one of these seat numbers will be circled and that will be your assigned seat for the first test. Try to arrive early to obtain your assigned seat. Unless you happen to sit at the end of a row, you will be required to stay for the full 50 minute testing period. The TAs will not answer any questions during the test.

Be sure to go to both discussion sections this week.

Feb 15 (Wed) Test 1 (given during lecture)
Feb 16 (Thu) Discussion sections still meet today. Your homework is to read section 3.1 and do #3–30, 33, 35, 47, 51, 53.
Feb 17 (Fri) Read sections 3.1–3.2 to learn the short-cut methods for finding derivatives. In section 3.1 do #3–30, 33, 35, 47, 51, 53. In section 3.2 do the odd problems from #3–33. There will be a quiz Thursday on sections 3.1–3.2.

We discussed the results of the first test along with strategies for improving one's score on future tests. Students who wish to drop our 5 credit hour course, have the option of adding a second eight week course.

We saw from the graphical interpretation of a derivative as a slope that the derivative of a constant is 0 since the graph of a constant function is a horizontal line which has slope 0. We also used this approach to see that the derivative of f(x) = mx + b is f ′(x) = m. In addition to the graphical approach, thinking about the derivative as a rate of change gives the same result. More formally, we use limits to prove these and the following derivative rules.

• ( c ) ′ = 0     (Do a proof with limits on your own)
• ( mx + b ) ′ = m     (Do a proof with limits on your own)
• ( xn ) ′ = nxn−1     (True for all real numbers n, our proof with limits was only for positive integers n)
• ( cf(x) ) ′ = cf ′(x)     (Do a proof with limits on your own)
• ( f(x) + g(x) ) ′ = f ′(x) + g ′(x)     (A proof with limits was given in lecture)
• ( f(x) − g(x) ) ′ = f ′(x) − g ′(x)     (Do a proof with limits on your own)
• ( f(x)g(x) ) ′ = f ′(x)g(x) + f(x)g ′(x)     (A proof with limits was given in lecture)
• ( f(x) / g(x) ) ′ = ( f ′(x)g(x) − f(x)g ′(x) ) / ( g(x) )2     (Try a proof with limits on your own)
These last two rules are referred to as the Product Rule and the Quotient Rule, respectively.

Next week we will use limits to derive short-cut methods for finding derivatives of all the basic functions. We will also discuss the Chain Rule for the derivative of a composition of functions. Here are the basic derivative rules I plan to discuss. It will be helpful for you to quickly memorize (or be able to derive) these rules. I plan to give more frequent quizzes to check that you are keeping up.

• ( c ) ′ = 0     (c is any constant)
• ( xn ) ′ = nxn−1
• ( ex ) ′ = ex
• ( ax ) ′ = ax ln(a)     (a > 0)
• ( ln(x) ) ′ = 1 / x
• ( sin(x) ) ′ = cos(x)
• ( cos(x) ) ′ = − sin(x)
• ( tan(x) ) ′ = sec2(x)
• ( cot(x) ) ′ = − csc2(x)
• ( sec(x) ) ′ = sec(x)tan(x)
• ( csc(x) ) ′ = − csc(x)cot(x)
• ( tan−1(x) ) ′ = 1 / (1 + x2)
• ( sin−1(x) ) ′ = 1 / sqrt(1 − x2)
• ( cos−1(x) ) ′ = −1 / sqrt(1 − x2)
• ( sec−1(x) ) ′ = 1 / (x*sqrt(x2 − 1))
Week 6
Feb 20 (Mon) Read sections 3.3 and 3.4. In section 3.3 do the odd problems from #1–23. Read the notes below for a few additional homework problems on deriving the derivative formulas for all trigonometric functions. There will be a quiz Thursday on sections 3.1 and 3.2.

Although one can directly find the derivative of a quotient using limits, it is easier to rewrite w(x) = f(x) / g(x) as w(x)*g(x) = f(x) and take the derivative of both sides using the product rule. Solving for w ′(x) gives us a simpler proof of the quotient rule. If you search YouTube for quotient rule or quotient rule song you will likely find many different mnemonic devices or songs to remember this rule.

We have shown earlier that limθ → 0 sin(θ) / θ = 1. Another useful limit is limθ → 0 (cos(θ) − 1) / θ = 0. We proved this result by first multiplying numerator and denominator by cos(θ) + 1.

Recall the trigonometric identities found in Appendix D.

• sin(x + y) = sin(x) cos(y) + cos(x) sin(y)
• cos(x + y) = cos(x) cos(y) − sin(x) sin(y)
Now from the definition of the derivative as a limit we obtain ( sin(x) ) ′ = limh → 0 ( sin(x + h) − sin(x) ) / h. Expanding sin(x + h) with the first identity above and rearranging terms we proved that ( sin(x) ) ′ = cos(x). For homework, students should use a similar technique to prove that ( cos(x) ) ′ = − sin(x).

Writing tan(x) = sin(x) / cos(x) we used the quotient rule to prove that ( tan(x) ) ′ = sec2(x). Writing sec(x) = 1 / cos(x) we used the quotient rule to prove that ( sec(x) ) ′ = sec(x)tan(x). For homework, students should rewrite cot(x) and csc(x) in terms of sin(x) and cos(x) to obtain the derivative of each of these functions.

Using limits the derivative of f(x) = ax is f ′(x) = limh → 0 ( (ax + h − ax) / h ) = limh → 0 ( (ax ah − ax) / h ) = ax limh → 0 ( (ah − 1) / h ) = ax f ′ (0). Thus the derivative of ax is ax multiplied by the slope of the curve y = ax at x = 0. This slope turns out to be ln(a) so that ( ax ) ′ = ax ln(a). In particular we get ( ex ) ′ = ex.

We can use the product rule to show that ( f(x) g(x) h(x) ) ′ = f ′(x) g(x) h(x) + f(x) g ′(x) h(x) + f(x) g(x) h ′(x). This generalizes naturally to the product of more terms.

Feb 22 (Wed) Read section 3.4. In section 3.4 do the odd problems from #7–55. Tomorrow's quiz is on sections 3.1 and 3.2.

The following table summarizes the derivative notation used in our textbook for first semester calculus. The prime notation is due to Joseph Louis Lagrange and Leibniz notation is due to Gottfried Wilhelm Leibniz. You should be comfortable using any of these notations for derivatives. For tomorrow's quiz you will be asked to correctly use Leibniz notation when computing derivatives. When using derivatives in other courses it may be worthwhile to see http://en.wikipedia.org/wiki/Notation_for_differentiation for other derivative notation due to Euler and Newton.

Derivative Notation
prime notation Leibniz notation
function f(x) = x3 y = x3 x3 y = x3 x3
1st derivative f ′(x) = 3x2 y ′ = 3x2 (x3) ′ = 3x2 dy/dx = 3x2 d/dx (x3) = 3x2
2nd derivative f ′′(x) = 6x y ′′ = 6x (x3) ′′ = 6x d2y/dx2 = 6x d2/dx2 (x3) = 6x
3rd derivative f ′′′(x) = 6 y ′′′ = 6 (x3) ′′′ = 6 d3y/dx3 = 6 d3/dx3 (x3) = 6

Using Leibniz notation the derivative of P = t3 is dP/dt = 3t2 and the derivative of w = r2 is dw/dr = 2r. Use the variables given in the problem instead of always using y and x.

If you wish to evaluate the derivative of x3 at 5 we have the following notation.

• Prime notation: f(x) = x3 ⇒ f ′(x) = 3x2 ⇒ f ′(5) = 75.
• Leibniz notation: y = x3 ⇒ dy/dx = 3x2 ⇒ dy/dx |x=5 = 75.
We used the product rule to prove that ( f(x) g(x) h(x) ) ′ = f ′(x) g(x) h(x) + f(x) g ′(x) h(x) + f(x) g(x) h ′(x). This generalizes naturally to the product of more terms.

Since ( f(x) )2 = f(x) f(x), ( f(x) )3 = f(x) f(x) f(x), etc., I used this generalized product rule to obtain the following derivatives.

• ( f(x)2 ) ′ = 2 f(x) f ′(x)
• ( f(x)3 ) ′ = 3 ( f(x) )2 f ′(x)
• ( f(x)4 ) ′ = 4 ( f(x) )3 f ′(x)
This generalizes to the following derivative rule.
• ( f(x)n ) ′ = n ( f(x) )n−1 f ′(x)
Our proof works when n is a positive integer but this rule is in fact true for all real numbers n. This is a special case of the more general chain rule for derivatives of the composition of functions.

Chain Rule: ( f(g(x)) ) ′ = f ′(g(x)) g ′(x)

The chain rule can also be written as dy/dx = (dy/du) * (du/dx) but we won't talk about this approach until next lecture.

Using the chain rule we found derivatives for the following functions.

• y = (x5 + 4x2 + 3)10
• y = tan5(x)
• y = e3x
• y = 1 / (x3 + 2)5
• y = ( 4sin(x) + 2cos(x) )1/2
• y = cot(x4 + 8x)
• y = esec(x)
• w = ( sin( et3 + 4t ) )1/2
• y = e3ln( e2ln(x) )
• P = ( x2 / (x5 + 1) )4
We proved that ( ln(x) ) ′ = 1/x in two different ways. You can just memorize the derivative formula but it is still useful to know the techniques used in these proofs.
1. Rewrite y = ln(x) as x = ey. Obtain dx/dy = ey. Now dy/dx = 1 / (dx/dy) = 1 / ey = 1 / x.
2. Note that eln(x) = x. Take the derivative with respect to x of both sides of this equation to obtain eln(x) * ( ln(x) ) ′ = 1. Now ( ln(x) ) ′ = 1 / eln(x) = 1 / x.
Feb 23 (Thu) Quiz #4 on sections 3.1 and 3.2 will be given during today's discussion section.
Feb 24 (Fri) Read sections 3.5 and 3.6. In section 3.5 do #5, 7, 9, 11, 13, 15, 17, 19, 29, 30, 31, 32, 49, 50, 51, 57. In section 3.6 do #3, 5, 6, 11, 13, 19, 31, 34, 39, 43, 45.

Here is one method for determining the derivative of sin−1(x).

• sin(sin−1(x)) = x
• (sin(sin−1(x))) ′ = (x) ′
• cos(sin−1(x))*( sin−1(x) ) ′ = 1   (Note the use of the chain rule.)
• ( sin−1(x) ) ′ = 1 / cos(sin−1(x))
• ( sin−1(x) ) ′ = 1 / √(1 − x2)   (The techniques from lecture on September 12th are used to show that cos(sin−1(x)) = √(1 − x2).)
For additional homework use a similar approach to obtain the following derivative rules.
• ( tan−1(x) ) ′ = 1 / (1 + x2)
• ( sec−1(x) ) ′ = 1 / (x * √(x2 − 1))
From the change of base formula we know that logb(x) = ln(x) / ln(b). Thus ( logb(x) ) ′ = ( ln(x) / ln(b) ) ′ = 1 / (x ln(b)). Although you can memorize this derivative formula, it is probably best just to use the change of base formula and then differentiate.

I discussed logarithmic differentiation to obtain derivatives of the following functions.

Example 1

• y = ln( x5 ex3 (x4 + 3)6 )
• y = ln( x5 ) + ln( ex3 ) + ln( (x4 + 3)6 )
• y = 5 ln(x) + x3 + 6 ln(x4 + 3)
• y ′ = 5/x + 3x2 + 6*4x3/(x4 + 3)
Example 2
• y = x3 * (x4 + 3)1/2 / (5x + 1)8
• ln(y) = ln( x3 * (x4 + 3)1/2 / (5x + 1)8 )
• ln(y) = ln( x3 ) + ln( (x4 + 3)1/2 ) − ln( (5x + 1)8 )
• ln(y) = 3 ln(x) + 1/2 ln(x4 + 3) − 8 ln(5x + 1)
• d/dx ( ln(y) ) = d/dx ( 3 ln(x) + 1/2 ln(x4 + 3) − 8 ln(5x + 1) )
• (1/y)*y ′ = 3/x + 1/2*4x3/(x4 + 3) − 8*5/(5x + 1)   (We used the chain rule on the left-hand side of the equation.)
• y ′ = y * ( 3/x + 2x3/(x4 + 3) − 40/(5x + 1) )
• y ′ = x3 * (x4 + 3)1/2 / (5x + 1)8 * ( 3/x + 2x3/(x4 + 3) − 40/(5x + 1) )
Example 3
• y = xx
• ln(y) = ln( xx )
• ln(y) = x*ln(x)
• d/dx ( ln(y) ) = d/dx ( x*ln(x) )
• (1/y)*y ′ = (1)*ln(x) + (x)*(1/x) (We used the chain rule on the left-hand side of the equation and the product rule on the right-hand side of the equation.)
• (1/y)*y ′ = ln(x) + 1
• y ′ = y * ( ln(x) + 1 )
• y ′ = xx * ( ln(x) + 1 )
For additional homework use this technique of logarithmic differentiation to determine the derivative of y = xcos(x).

We found the slope of the tangent line to the curve x2 + y2 = 25 at the point (3, 4) in two ways.

• Method 1: We solved for y explicitly to get y = √( 25 − x2 ). Note that at the point (3, 4) we use the positive square root. Using the chain rule we found y ′ = −x / √(25 − x2). We plugged in x=3 to obtain that the slope of the tangent line at the point (3, 4) is −3/4.
• Method 2:
• x2 + y2 = 25
• d/dx (x2 + y2) = d/dx (25)
• 2x + 2y*y ′ = 0
• y ′ = −x / y
• At (x, y) = (3, 4), y ′ = −3 / 4
This 2nd method is called implicit differentiation and you have to think carefully to see why we used the chain rule to obtain that the derivative with respect to x of y2 is 2y*y ′.

We found the derivative dy/dx for the following implicitly defined functions. For the first example we also plugged in (x, y) = (2, 4) to find the slope of the curve at that point.

• x3 + y3 = 9xy   (This curve is called the folium of Descartes. Enter plot(x^3 + y^3 = 9xy) at Wolfram Alpha to see a graph of this curve.)
• y2 + x5y + 2x = 0
For the second example above, we could have used the quadratic formula to obtain that y = ( −x5 ± sqrt(x10 − 8x) ) / 2 and then found the derivative, but implicit differentiation is an easier approach.
Week 7
Feb 27 (Mon) Read sections 3.7 and 3.8. In section 3.7 do #7, 8, 9, 10. In section 3.8 do #3, 4, 8, 9, 10, 11, 12. There will be a quiz Thursday on sections 3.3, 3.4 and 3.5. I have cancelled today's office hours, but will be available Tuesday 3-5:30pm in 121/123 Altgeld Hall.

We found the derivative of xcos(x) using logarithmic differentiation as follows.

• y = xcos(x)
• ln(y) = ln( xcos(x) )
• ln(y) = cos(x)*ln(x)
• d/dx ( ln(y) ) = d/dx ( cos(x)*ln(x) )
• (1/y)*y ′ = −sin(x)*ln(x) + cos(x)/x   (We used the chain rule on the left-hand side of the equation and the product rule on the right-hand side of the equation.)
• y ′ = y * ( −sin(x)*ln(x) + cos(x)/x )
• y ′ = xcos(x) * ( −sin(x)*ln(x) + cos(x)/x )
Using logarithmic differentiation I proved that ( xn ) ′ = n xn−1 is valid for all real numbers n. Our earlier proof only worked for positive integers n.

See how the chain rule is used for each of the following problems.

• d/dx ( sin(x4 + 5x) ) = cos(x4 + 5x) * d/dx (x4 + 5x) = cos(x4 + 5x) * (4x3 + 5)
• d/dx ( sin(tan(x)) ) = cos(tan(x)) * d/dx (tan(x)) = cos(tan(x)) * sec2(x)
• d/dx ( sin(♣) ) = cos(♣) * d/dx (♣)   (The symbol ♣ could be replaced with any function.)
• d/dx ( sin(y) ) = cos(y) * d/dx (y) = cos(y) * dy/dx   (This follows the same form as each example above.)
In the last example above we used the chain rule in the same way we used it for the other examples, but students too often miss this since we didn't explicitly write y as a function of x.

I solved a couple of problems from the homework in sections 3.5 and 3.6.

My only applications from section 3.7 are the ones concerning position, velocity and acceleration. Since I have already discussed these concepts, I will expect students to already have the tools needed to solve these homework problems.

The following are examples of differential equations (equations which include derivatives).

• y ′′ + 3y ′ + y = x2
• ( dy/dx )2 = 4y + 3
Engineering majors and some other majors require a course in differential equations. They will need to analyze a differential equation by finding a formula, a graph, or numerical estimates for a function y which satisfies the differential equation.

We found formulas for y as a function of x to satisfy each of the following differential equations along with the given initial values.

• dy/dx = 3ex, y(0) = 8   (Solution: y = 3ex + 5)
• dy/dx = 10x, y(2) = 30   (Solution: y = 5x2 + 10)
• dy/dx = 8x, y(0) = 20   (Solution: y = 4x2 + 20)
• dy/dx = 8y, y(0) = 20   (Solution: y = 20e8x)
For the first three examples here students came up with the correct solution by thinking about the process of differentiation in reverse. That is, given the derivative find the function. Although the last two examples look very similar, their solutions are not at all similar. Carefully check the varia bles used to see why they lead to two different solutions. For dy/dx = 8y we needed to find a function whose derivative is the same function multiplied by 8. After some trial and error we discovered that the function y = 20e8x worked.

We obtain in general that the differential equation dy/dx = ky where k is a constant has solution y = Cekx. If an initial value is given then we can plug in that particular point to solve for unknowns like C.

I did not have time for the following examples in lecture, but it is hopefully clear from our discussion why we obtain the given solutions.

• dy/dx = 6x, y(0) = 10   (Solution: y = 3x2 + C. By plugging in (x, y) = (0, 10) we find that C = 10 so that our solution becomes y = 3x2 + 10.)
• dy/dx = 12x, y(2) = 8   (Solution: y = 6x2 + C. By plugging in (x, y) = (2, 8) we find that C = −16 so that our solution becomes y = 6x2 − 16.)
• dy/dx = 4y, y(0) = 5   (Solution: y = Ce4x. By plugging in (x, y) = (0, 5) we find that C = 5 so that our solution becomes y = 5e4x.)
• dy/dx = 3y, y(5) = 2   (Solution: y = Ce3x. By plugging in (x, y) = (5, 2) we find that C = 2/e15 so that our solution becomes y = (2/e15)e3x or y = 2e3x−15.)
Feb 29 (Wed) Read section 3.9. In section 3.9 do #6, 10, 13, 15, 20, 22, 24, 27, 28, 30, 31, 38, 41. Prepare for tomorrow's quiz on sections 3.3, 3.4 and 3.5. There will be two quizzes next week. The test is in two weeks so don't fall behind. Sections 3.9 and 4.7 include a lot of word problems so be prepared to work hard if this is an area of weakness for you.

I solved the first two problems on this related rates worksheet. Do the third problem on your own. Solutions will be provided soon.

Continuing our discussion of differential equations and exponential functions...

If dy/dx is given in terms of x (the independent variable), then we must think about the process of differentiation in reverse in order to determine a formula for y as a function of x. By adding an arbitrary constant C, we obtain a family of functions which have the given formula for dy/dx. If an initial value is given then we can plug this point into the formula for y in order to determine the value of C. Since we often use variables other than x or y, you must pay close attention to the variables before determining a solution to the differential equation. We have the following examples.

• dy/dx = 15x2, y(2) = 30   (Solution: y = 5x3 + C. By plugging in (x, y) = (2, 30) we find that C = −10 so that our solution is y = 5x3 − 10.)
• dw/dp = 40p4, w(−1) = 10   (Solution: w = 8p5 + C. By plugging in (p, w) = (−1, 10) we find that C = 18 so that our solution is w = 8p5 + 18.)
• dh/dv = 6v, h(0) = 4   (Solution: h = 3v2 + C. By plugging in (v, h) = (0, 4) we find that C = 4 so that our solution is h = 3v2 + 4.)
• dP/dt = 12t, P(1) = 25   (Solution: P = 6t2 + C. By plugging in (t, P) = (1, 25) we find that C = 19 so that our solution is P = 6t2 + 19.)
If dy/dx is given in terms of y (the dependent variable), then determining a formula for y as a function of x can be more difficult. We saw in general that the differential equation dy/dx = ky where k is a constant has solution y = Cekx. Note that this differential equation is different from dy/dx = kx. If an initial value is given then we can plug this point into the formula for y in order to determine the value of C. Since we often use variables other than x or y, you must pay close attention to the variables before determining a solution to the differential equation. We have the following examples. Compare them to the last two examples above.
• dh/dv = 6h, h(0) = 4   (Solution: h = Ce6v. By plugging in (v, h) = (0, 4) we find that C = 4 so that our solution is h = 4e6v.)
• dP/dt = 12P, P(1) = 25   (Solution: P = Ce12t. By plugging in (t, P) = (1, 25) we find that C = 25/e12 so that our solution is P = (25/e12)e12t or P = 25e12t−12.)
We looked at two populations. One population is currently 100 and increasing by 5 people per year. Another population is currently 100000 and increasing by 5000 people per year. These two populations are growing at the same relative growth rate 5%. Using P for population at time t, we see that both situations lead to the differential equation dP/dt = 0.05P. This is one example where differential equations come up naturally in the real world.

If a population is currently 200 and growing at a constant relative growth rate of 3%, then this leads directly to the differential equation dP/dt = 0.03P with P(0) = 200. Solving this differential equation gives us the following formula for the population: P = 200e0.03t.

If a quantity A is proportional to B, then this means that A = k*B where k is a constant. That is, you can translate "is proportional to" to "equals a constant times". Since A = π r2 gives the area of a circle, we see that the area is proportional to the square of the radius. Since V = 4/3*π r3 gives the volume of a sphere, we see that the volume is proportional the cube of the radius. If a population is growing at a rate which is proportional to the population size, this translates to dP/dt = k*P which has solution P = Cekt.

Students should know the meaning of the term half-life and be able to use exponential functions to help solve problems involving half-lives. Although y = C*ax and y = C*ekx are both valid formulas for exponential functions, we will begin using the second form more often. The example solved in class was to determine how long it takes for 100mg of caffeine in the bloodstream after a cup of coffee to be reduced to 10% of that amount. The half-life of caffeine in the bloodstream is about 4 hours for most people but closer to 10 hours for pregnant women.

Mar 1 (Thu) Quiz #5 on sections 3.3, 3.4 and 3.5 will be given during today's discussion section.
Mar 2 (Fri) For homework read section 4.1 and do #16–25, 30, 41, 43, 49–60, 63, 75 from that section. Quiz #6 on sections 3.6, 3.7 and 3.8 will be given during Tuesday's discussion section. Quiz #7 will be a take-home quiz on sections 3.9, 4.1, 4.3, 4.4 and 4.7. It will be due in lecture next Friday.

In lecture today I further discussed strategies for solving related rates problems. I solved problem #41 from section 3.9 and the third problem on the related rates worksheet.

We obtained a graph of the basic shape of f(x) = x4 − 4x3 + 16x − 16 by first looking at its derivative f ′(x) = 4x3 − 12x2 + 16. Since the derivative factors as f ′(x) = 4(x + 1)(x − 2)2 we can quickly see which x-values cause the derivative to be positive, negative or zero. This tells us where the graph of f(x) is increasing, decreasing or level.

We used this same approach to obtain a graph of f(x) = 5xe−2x = 5x / e2x. We found f ′(x) = (5 − 10x) / e2x and noted that f ′(x) > 0 for x < 1/2, f ′(x) = 0 for x = 1/2, and f ′(x) < 0 for x > 1/2. Thus the graph of f(x) is increasing for x < 1/2, level at x = 1/2, and decreasing for x > 1/2. Even though the graph of f(x) is decreasing for x > 1/2, we see from the formula for f(x) that the y-values never become negative. We used this along with limx → ∞ f(x) = limx → ∞ ( 5x / e2x ) = 0 to get a better graph for this function.

The derivative function f ′(x) tells us the shape of the graph of f(x) but not the y-values. When actually graphing a function f(x) we should plug specific x-values into f(x) to obtain the corresponding y-values. Next week we'll see how the second derivative gives us further information about the shape of a graph.

I introduced terms such as critical numbers, absolute maximum, absolute minimum, local maximum and local minimum.

I discussed the Extreme Value Theorem and the Closed Interval Method.

Week 8 (Deadline to drop this course without a grade of W is Friday, March 9, 2012)
Mar 5 (Mon) Read sections 4.3 and 4.7. In section 4.3 do #10, 13, 17, 18, 33, 39, 43, 46, 48, 53, 86. In section 4.7 do #5, 6, 13, 14, 19, 21, 32, 34, 35, 38, 49, 54. Quiz #6 on sections 3.6, 3.7 and 3.8 will be given during Tuesday's discussion section. Quiz #7 will be a take-home quiz on sections 3.9, 4.1, 4.3, 4.4 and 4.7. It will be distributed in lecture on Wednesday and due in lecture on Friday.

We used the Closed Interval Method to determine the absolute minimum and absolute maximum values for the function f(x) = x3 − 6x2 + 5 on the interval [−3, 5].

On an interval we have the following.

• f ′ > 0 ⇒ f is increasing
• f ′ < 0 ⇒ f is decreasing
• f ′′ > 0 ⇒ f ′ is increasing ⇒ f is concave up
• f ′′ < 0 ⇒ f ′ is decreasing ⇒ f is concave down
In addition to looking at whether f ′ is increasing or decreasing to determine the concavity of f, I also discussed the book's definition of concavity in terms of the tangent lines to a curve.

We used information about the first and second derivative to obtain a graph of f(x) = 2x3 + 3x2 − 36x.

A rectangle is to be inscribed in a semi-circle of radius 2. What is the largest possible area and what are the dimensions that will give this area? For our solution we drew the upper half of the circle of radius 2 centered at the origin. We noted that y = √(4 − x2) for each point on this semi-circle. If we use (x, y) as the coordinates of the point at the upper right corner of the inscribed rectangle, then the area of the rectangle is A = base*height = 2x*y = 2x*√(4 − x2). The next step is to maximize this area on the interval [0, 2] using the Closed Interval Method. That is, using the formula A = 2x*√(4 − x2), you will plug in the endpoints x=0 and x=2 as well as any points on the interval (0, 2) for which A ′ = 0 or A ′ does not exist.

Mar 6 (Tue) Quiz #6 on sections 3.6, 3.7 and 3.8 will be given during today's discussion section.
Mar 7 (Wed) Read section 4.4. In 4.4 do #7, 11, 17, 18, 19, 21, 25, 33, 41, 45, 49, 50, 55, 57, 61, 62, 67. Next Wednesday's test will cover sections 3.1–3.9, 4.1, 4.3, 4.4, 4.7. Quiz #7 is a take-home quiz on sections 3.9, 4.1, 4.3, 4.4 and 4.7. It will be distributed on Friday and should be turned in at the beginning of Monday's lecture.

I briefly discussed inflection points, the first derivative test and the second derivative test.

We used l'Hospital's Rule to determine the following limits.

1. limx → 0 ( (3x − sin(x)) / x ) = 2
2. limx → 0 ( (sqrt(1 + x) − 1 − x/2) / x2 ) = −1/8
3. limx → 0 ( (1 − cos(x)) / (x + x2) ) = 0
4. limx → ∞ ( ln(x) / (2sqrt(x)) ) = 0
5. limx → ∞ ( e2x / (x2 + 3x + 5) ) = ∞
6. limx → ∞ ( x2 / ex ) = 0
7. limx → 0+ ( x ln(x) ) = 0
8. limx → 0 ( 1 / sin(x) − 1 / x ) = 0   (We began this problem – finish it for homework.)
9. limx → ∞ ( (1 + 1 / x)x ) = e   (We began this problem – finish it for homework.)
For examples (4), (5) and (6) one should see how orders of growth to infinity should immediately give us answers without the need for l'Hospital's Rule. Specifically as x approaches ∞ the following functions approach ∞ from slowly to quickly in the order shown.

(slowly)   ln(x), ..., x1/3, x1/2, x, x2, x3, ..., ex   (quickly)

If you need to take the limit of the ratio of two such functions, then the slowness or quickness of growth toward ∞ should be enough to tell you if the limit of the ratio is 0 or ∞. For example one should immediately see that limx → ∞ ( (4x1000 + 5x50 + 10) / (0.001e2x) ) = 0 since the numerator approaches ∞ slowly while the denominator approaches ∞ quickly.

The following are considered indeterminate forms so it is helpful to have a technique such as l'Hospital's Rule for determining limits in these cases.

• 0 ÷ 0
• ±∞ ÷ ±∞
• 0 × ±∞
• ∞ − ∞
• 1
• 00
• 0
You should think about why forms like 0, ∞1 and 1 ÷ 0 are not indeterminate. For example, we can directly obtain limx → ∞ ( 1 / x )x2 = 0 and limx → 0 ( cos(x) / x2 ) = ∞.
Mar 8 (Thu)
Mar 9 (Fri) Quiz #7 is a take-home quiz on sections 3.9, 4.1, 4.3, 4.4 and 4.7. It should be turned in to Mr. Murphy at the beginning of Monday's lecture.

Use the Test 2 Notes to begin your preparation for Wednesday's test.

Mar 9 (Fri)
Week 9
Mar 12 (Mon)
Mar 14 (Wed) Test 2 (given during lecture)
Mar 15 (Thu) Discussion sections still meet today. Your homework is to read section 4.9 and do #1–17, 20–22, 25–33, 41–43, 65, 69, 73–75.
Mar 16 (Fri) Finish the homework from section 4.9. Read section 5.1 very carefully. Do #3, 4, 13, 14, 15, 18, 20 from section 5.1.

A function F is called an antiderivative of f on an interval I if F ′(x) = f(x) for all x in I. Since you have already memorized a lot of short-cut derivative rules, you can use this knowledge to quickly determine the following antidervatives.

Function Particular antiderivative
xn (n ≠ −1) xn + 1 / (n + 1)
x−1 = 1/x ln |x|
ex ex
cos(x) sin(x)
sin(x) −cos(x)
sec2(x) tan(x)
csc2(x) −cot(x)
sec(x) tan(x) sec(x)
csc(x) cot(x) −csc(x)
1 / (1 + x2) tan−1(x)
1 / √(1 − x2) sin−1(x)

It is straightforward to find an antiderivative for a constant multiplied by a function or the sum of two or more functions. From the more complicated short-cut derivative rules for products, quotients and the composition of functions, one should expect that it is not as straightforward to find antiderivatives for products, quotients and the composition of functions. We will discuss some of these later. For now if you see a complicated function, you should try to rewrite the expression using algebra or trigonometry before finding an antiderivative.

In particular we made the following simplifications before finding antiderivatives for #14 and #22 in section 4.9.

• f(t) = (3t4 − t3 + 6t2) / t4 = 3t4 / t4 − t3 / t4 + 6t2 / t4 = 3 − t−1 + 6t−2
• f(x) = (2 + x2) / (1 + x2) = (1 + (1 + x2)) / (1 + x2) = 1 / (1 + x2) + 1
For the second problem above we also could have rewritten the expression using polynomial long division.

When the rate of change of a quantity is positive, we saw graphically that the total change in that quantity on some interval could be represented by the area between the function and the horizontal axis on that interval.

We looked at the simple problem where a car travels at a constant rate of 50 miles per hour for a 2-hour period. The total change in position is (50 miles/hour) × (2 hours) = 100 miles. If we graphed the horizontal line y = 50 and shaded in the area between the horizontal axis and this line on a 2-hour time interval, we obtain that the area of this rectangle is 50 × 2 = 100.

We next looked at an object which travels at v(t) = √(t) feet per second between times t = 1 and t = 3 seconds. By breaking the interval [1, 3] into smaller subintervals we could approximate the total change in position on each subinterval as the area of a rectangle. The total change in position would then be approximated by the sum of the areas of these rectangles. Using 4 subintervals of equal width and a right Riemann sum, we obtained the following approximation.

Total change in position ≈ √(1.5) × 0.5 + √(2) × 0.5 + √(2.5) × 0.5 + √(3) × 0.5 ≈ 2.98.

Graphically we see why the above approximation is an overestimate to the actual change in position. We discussed how one would obtain an underestimate. We also discussed the idea that using more subintervals usually leads to a better approximation. You may need a calculator for some of the homework problems where the computations are lengthy.

I used a few examples to introduce Σ notation for sums and relayed the story of Gauss quickly obtaining the sum of 1 + 2 + 3 + ... + 100 as a young boy. More generally we find that 1 + 2 + 3 + ... + n = n(n + 1)/2.

Week 10 (Spring Break!)
Week 11
Mar 26 (Mon)
Mar 28 (Wed) Read section 5.2 very carefully. Start working on #2, 11, 18, 21, 22, 29, 33, 36, 37, 41, 48, 49, 52, 53, 55, 57, 59 from section 5.2. Quiz #8 on sections 4.9 and 5.1 will be given on Thursday.

Using Σ notation I wrote the sum 52 + 62 + 72 + ... + 202 in four different ways.

• Sum from k = 5 to 20 of k2
• Sum from i = 5 to 20 of i2
• Sum from j = 0 to 15 of (j + 5)2
• Sum from k = 8 to 23 of (k − 3)2
Using Σ notation we discussed the following sums.
• 1 + 2 + 3 + ... + (n − 1) + n = n(n + 1)/2
• 12 + 22 + 32 + ... + (n − 1)2 + n2 = n(n + 1)(2n + 1)/6
• 13 + 23 + 33 + ... + (n − 1)3 + n3 = [ n(n + 1)/2 ]2
Rewriting the terms in reverse order resulted in a proof of the formula for the first sum above. Try the same method for the other two sums. If this technique does not lead to a proof of the formulas given, can you think of another way to prove that the formulas are valid? I will discuss this Wednesday or Friday. For now you can simply memorize these formulas.

Using the second sum above, we found the sum from k = 1 to n of 5k2/n3 to be (5/n3)*n(n+1)(2n+1)/6. We then found the limit of this sum as n goes to infinity to be 10/6 = 5/3.

We looked at area as a limit of Riemann sums. I talked about right Riemann sums, left Riemann sums, midpoint Riemann sums, and sums where an arbitrary xk* was chosen on each interval [ xk−1, xk ] in order to generate f(xk*). We then used this limit approach with right Riemann sums to evaluate the area between the x-axis and f(x) = 2x on the interval [1, 5]. Of course since the shape is just a trapezoid you can use geometry to find our answer more simply and should compare the two answers.

For another example we calculated the limit of right Riemann sums in order to determine the area between the x-axis and f(x) = x2 on the interval [2, 8]. This would be more difficult to check using geometry but you should at least have an approximate answer in mind to compare to the exact answer obtained in lecture.

Mar 29 (Thu) Quiz #8 on sections 4.9 and 5.1 will be given during today's discussion section.
Mar 30 (Fri) Read section 5.3. In section 5.3 do #23, 24, 28, 31, 32, 33, 35, 39, 45, 48, 51.

Using Σ notation we again looked at the following sums.

• 1 + 2 + 3 + ... + (n − 1) + n = n(n + 1)/2
• 12 + 22 + 32 + ... + (n − 1)2 + n2 = n(n + 1)(2n + 1)/6
• 13 + 23 + 33 + ... + (n − 1)3 + n3 = [ n(n + 1)/2 ]2
Students should know each of these sums. I showed a geometric argument for the sum of the first n positive integers. I indicated the beginning of a geometric argument for the sum of the first n squares. I proved the sum of the first n squares by looking at the collapsing sum from k = 1 to n of (k + 1)3 − k3 in two different ways. For the sum of the first n cubes look at the sum from k = 1 to n of (k + 1)4 − k4 in two different ways. This technique generalizes to higher powers.

I gave the definition found on page 372 for the definite integral including the term integrable for when the given limit exists.

Theorem: If f is continuous on [a, b] then f is integrable on [a, b].

We looked at the definite integral of 2x from 0 to 3. Since 2x is continuous on [0, 3] we know that it is integrable on [0, 3]. We can evaluate the definite integral in multiple ways.

1. We can determine the answer by using a limit of Riemann sums.
2. We can shade in the region under the graph of 2x on [0, 3] to see it is a triangle whose area we can compute using 1/2(base)*(height).
3. We can use antiderivatives as mentioned in class by some students.
We looked at the definite integral of ex2 from 0 to 3. Since ex2 is continuous on [0, 3] we know that it is integrable on [0, 3]. However we are more limited in the approaches we can use to determine this limit.
1. The limit of Riemann sums is too difficult to compute.
2. The region under the graph of ex2 on [0, 3] is not a simple shape whose area we know.
3. We cannot find a formula for an antiderivative of ex2.
Our best option at this point is to approximate the definite integral of ex2 from 0 to 3 by using a Riemann sum with a fixed number of subintervals. If we don't mind the lengthy computations, then it is a good idea to use as many subintervals as possible, but I only used about 4 in lecture.

Calculators have approximation techniques for evaluating definite integrals. For example the definite integral of 15x4 from x = 1 to 3 can be approximated in the following ways.

• On a TI-83 or TI-84 graphing calculator, enter fnInt(15x^4, x, 1, 3). The fnInt command is found as a submenu of the MATH button.
• On a TI-89 graphing calculator, select F3 from the HOME screen followed by 2 to integrate. This should give you the integral symbol along with the left parenthesis. You should complete this line so that you see the integral symbol followed by (15x^4, x, 1, 3).
• At Wolfram Alpha, enter int 15x^4 from 1 to 3. Wolfram Alpha is very flexible so there are many other ways to enter this integral.
I discussed further the approach of using antiderivatives to determine the value of definite integrals. The following application should suggest why we expect the method to be valid.

Suppose that a population is expected to increase at a rate of 6t + 2 people per year where t represents the number of years from now. What is the expected change in population between years t = 1 and t = 3? We began to solve this in two ways.

Method 1: After graphing 6t + 2 on the interval [1, 3], we saw that a limit of Riemann sums would give the exact change in population. Since by definition this limit is the definite integral, we saw that the exact change in population is the definite integral of 6t + 2 as t goes from 1 to 3.

Method 2: If we let P(t) represent the population at time t, then P ′(t) represents the rate of change of population at time t. Thus P ′(t) = 6t + 2. By solving this differential equation we obtain the formula P(t) = 3t2 + 2t + C as the population at time t. The exact change in population is then seen to be P(3) − P(1) = (33 + C) − (5 + C) = 28 people. Notice that C drops out here.

Our second method for this problem demonstrates the Net Change Theorem found on page 401 which basically says that the definite integral of a rate of change gives the total change.

More generally we have the Fundamental Theorem of Calculus (part 2) from section 5.3 which shows a wonderful shortcut for evaluating many definite integrals. We will discuss this more fully next week.

Week 12
Apr 2 (Mon) Read section 5.4. In section 5.4 do #3, 5, 6, 15, 16, 17, 18, 27, 31, 37, 43, 53, 54, 64. Many of these are straightforward drill problems using antiderivatives instead of limits. If necessary you should do more than the assigned problems so that you become quick at obtaining these answers. Quiz #9 on sections 5.2, 5.3 and 5.4 will be given on Thursday.

For a given definite integral, I looked at how to write it as a limit of right Riemann sums as compared to a limit of left Riemann sums. Students should also think about how to write a definite integral as a limit of midpoint Riemann sums.

I solved #37 and #57 from the homework in section 5.2. For #37 we first wrote the definite integral as a limit of right Riemann sums. Since the limit appeared difficult to compute, we instead solved the problem geometrically by graphing the integrand and noting that the area of the region between the x-axis and the curve is simply the sum of the area of a rectangle and the area of a quarter circle. For #57 I used the minimum and maximum values of the function in order to approximate the definite integral of the function. Drawing a graph of the integrand and thinking about areas really helps here, but we are simply using the third comparison property listed below.

For integrable functions f(x) and g(x), I discussed the following comparison properties.

• If f(x) ≥ 0 for a ≤ x ≤ b, then the definite integral of f(x) from a to b ≥ 0.
• If f(x) ≥ g(x) for a ≤ x ≤ b, then the definite integral of f(x) from a to bthe definite integral of g(x) from a to b.
• If m ≤ f(x) ≤ M for a ≤ x ≤ b, then m(b − a) ≤ the definite integral of f(x) from a to b ≤ M(b − a).
Suppose a population is currently 2000 and is expected to grow by 0.3t2 people per year where t represents the number of years from now. What is the expected change in population over the next 10 years? We solved this in two ways.

Method 1: We graphed 0.3t2 on the interval [0, 10] and then took the limit of Riemann sums to get the exact change in population. Since by definition this limit is the definite integral, we saw that the exact change in population is the definite integral of 0.3t2 as t goes from 0 to 10. When we computed this definite integral as a limit of Riemann sums, we obtained an answer of 100 people as the change in population.

Method 2: If we let P(t) represent the population at time t, then P ′(t) represents the rate of change of population at time t. Thus P ′(t) = 0.3t2 and P(0) = 2000. By solving this differential equation we obtain the formula P(t) = 0.1t3 + 2000 as the population at time t. The exact change in population is then seen to be P(10) − P(0) = 2100 − 2000 = 100 people.

Our second solution to this problem demonstrates the Net Change Theorem found on page 401 which basically says that the definite integral of a rate of change gives the total change.

More generally we have the Fundamental Theorem of Calculus (part 2) from section 5.3 which shows a wonderful shortcut for evaluating many definite integrals. We used this quick approach for the following problems.

I formally wrote out the Fundamental Theorem of Calculus (part 1) and the Fundamental Theorem of Calculus (part 2). Today we only discussed part 2 but students should read about part 1 in the book.

• Since sin(x) is an antiderivative of cos(x), the definite integral of cos(x) from 0 to π/2 = sin(π/2) − sin(0) = 1 − 0 = 1.
• Since 2x3 + x2 + ln|x| is an antiderivative of 6x2 + 2x + 1/x, the definite integral of 6x2 + 2x + 1/x from 1 to 2 = ( 2(2)3 + (2)2 + ln(2) ) − ( 2(1)3 + (1)2 + ln(1) ) = 17 + ln(2).
• Since tan(x) is an antiderivative of sec2(x), the definite integral of sec2(x) from 0 to π/4 = tan(π/4) − tan(0) = 1 − 0 = 1.
• The definite integral of x−2 from −2 to 1 = ???   (Students worked on this at their desks and came up with 6 or 7 different incorrect answers.)
Why did this last problem prove to be difficult for many of the students?

Recall that f(x) is integrable on an interval [a, b] if the appropriate limit of the Riemann sums exists (i.e. the definite integral of f(x) from a to b exists and is finite). Theorem 3 in section 5.2 states that if f is continuous on [a, b], then f is integrable on [a, b]. For the last example above, the integrand x−2 = 1/x2 is not continuous on the interval [−2, 1] and it turns out that the function is not integrable on the given interval. We should not attempt to directly use the Fundamental Theorem of Calculus for this definite integral. However I'm still not quite sure why students came up with such a large number of different incorrect answers. It must have been due to incorrect arithmetic or incorrect antiderivatives.

The indefinite integral represents the most general antiderivative of the integrand on an interval. Students should know the table of indefinite integrals found on page 398 but can temporarily ignore the last two where the integrand is sinh(x) or cosh(x).

Apr 4 (Wed) Read section 5.5 and do #8, 16, 17, 18, 20, 21, 22, 23, 25, 28, 32, 39, 40, 41, 44, 46, 48, 54, 57, 59, 60, 61, 65, 66, 67, 69, 81, 82 from that section. There will be a quiz Thursday on sections 5.2, 5.3 and 5.4.

We began class by looking at the difference between definite integrals and indefinite integrals.

What is the definite integral of f(x) from a to b? If f(x) is integrable on the interval [a, b], then the definite integral of f(x) from a to b is equal to a finite number. This finite number may be obtained using the definition of a definite integral as a limit of Riemann sums. However it is often easer to find this finite number using the Fundamental Theorem of Calculus (part 2) or simple geometry.

What is the indefinite integral of f(x)? The indefinite integral is the most general antiderivative of f(x) on an interval. Thus it is a family of functions which all have f(x) for a derivative.

Students should know the table of indefinite integrals found on page 398 but you can ignore the last two where the integrand is either of the hyperbolic functions sinh(x) or cosh(x).

We looked at the connection between definite integrals and area with the following examples.

• The definite integral from 0 to 5 of 6e2x. Since 6e2x > 0 on the interval [0, 5] we see that the value of each Riemann sum > 0 so we expect its limit, the definite integral, to be positive. We see that the definite integral represents the area of the region between the x-axis and the function on the given interval.
• The definite integral from π/2 to 3π/2 of cos(x). Since cos(x) ≤ 0 on the interval [π/2, 3π/2] we see that the value of each Riemann sum ≤ 0 so we expect its limit, the definite integral, to be ≤ 0. We see that the definite integral represents the negative of the area of the region between the x-axis and the function on the given interval.
More generally, for an integrable function f(x), to determine the definite integral of f(x) you can shade in the area between the graph of f(x) and the x-axis on the given interval. The definite integral represents the area shaded above the x-axis minus the area shaded below the x-axis. Area is positive but a definite integral can be positive or negative or zero.

I evaluated the following integrals from section 5.3.

• (#6) The indefinite integral of the square root of x3 + the cube root of x2 dx.
• (#18) The indefinite integral of sin(2x) / sin(x) dx.
• (#37) The definite integral from 0 to π/4 of (1 + cos2θ) / cos2θ dθ.
We looked at the following three indefinite integrals.

• The indefinite integral of 6x2(x3 + 4) dx.
• The indefinite integral of 6x2(x3 + 4)3 dx.
• The indefinite integral of 6x2(x3 + 4)10 dx.
For the first problem I rewrote the integrand as 6x5 + 24x2 before finding the most general antiderivative.

For the second problem I rewrote the integrand as 6x11 + 72x8 + 288x5 + 384x2 before finding the most general antiderivative.

Rewriting the integrand for the second problem was a little bit time-consuming. Rewriting the integrand for the third problem will be too time-consuming. We need a better way of handling problems like these.

In section 5.5 we see that the chain rule in reverse leads to the method of substitution. We used this method to evaluate the following integrals.

1. The indefinite integral of 6x2(x3 + 4)10 dx.
2. The indefinite integral of 1 / (3x + 10) dx.
3. The definite integral from 1 to 2 of 2x3(x4 + 5)8 dx.   (Change the limits of integration when using substitution.)
4. The definite integral from −1/3 to 1/3 of tan(x) dx.
For this last example we rewrote tan(x) as sin(x) / cos(x) before making the substitution u = cos(x). We did not finish the problem, but we'll discuss it more fully on Friday. You should think about the graph of tan(x) to see if you can anticipate the answer even before going through the details.
Apr 5 (Thu) Quiz #9 on sections 5.2, 5.3 and 5.4 will be given during today's discussion section.
Apr 6 (Fri) Read sections 6.1 and 4.2. In section 6.1 do #1, 8, 11, 12, 13, 17, 23, 25, 27, 29, 50, 51.

From the graph of tan(x) that we should expect the definite integral from −1/3 to 1/3 of tan(x) dx to be equal to 0. In fact if f(x) is an odd function integrable on [−a, a], we always get that the definite integral from −a to a of f(x) dx = 0.

We found the indefinite integral of sin(2x) dx by solving it three different ways.

1. Using the substitution u = 2x to obtain (−1/2)*cos(2x) + C
2. Writing sin(2x) = 2sin(x)cos(x) and using the substitution u = sin(x) to obtain sin2(x) + C
3. Writing sin(2x) = 2sin(x)cos(x) and using the substitution u = cos(x) to obtain −cos2(x) + C
These answers look different but are all correct. Through basic trigonometric identities we see that (−1/2)*cos(2x), sin2(x), and −cos2(x) simply differ by constants. Thus when adding an arbitrary constant, we arrive at the same family of functions.

Here are some quick ways to check that your evaluation of an indefinite integral is correct.

• Check by hand.
• Take the derivative of your result to make sure that you obtain the integrand. For example if you found that the indefinite integral of 15x4 dx is equal to 3x5 + C, you can guarantee your answer is correct by showing that (3x5 + C) ′ = 15x4.
• Check with technology.
• Use a TI-89 graphing calculator. To integrate 15x4 with respect to x, select F3 from the HOME screen followed by 2 to integrate. This should give you the integral symbol along with the left parenthesis. You should complete this line so that you see the integral symbol followed by (15x^4, x). Entering this command will give you the expected result of 3x5 but it leaves off the addition of an arbitrary constant C.
• Use Wolfram Alpha. To integrate 15x4 with respect to x, enter int 15x^4. Wolfram Alpha is very flexible so there are many other ways to enter this integral.
Here are some quick ways to check that your evaluation of a definite integral is correct.
• Check by hand.
• You could use the technique listed for indefinite integrals to make sure your antiderivative is correct before applying the Fundamental Theorem of Calculus.
• If the definite integral represents the area of some region, then estimate the area beforehand. I often draw a rectangle around the region and compute the area of this rectangle. I can see how much of the rectangle is taken up by the region and thus estimate the area of the region.
• Check with technology.
• Use a TI-83 or TI-84 graphing calculator. To integrate 15x4 from x = 1 to 3, enter fnInt(15x^4, x, 1, 3). The fnInt command is found as a submenu of the MATH button.
• Use a TI-89 graphing calculator. To integrate 15x4 from x = 1 to 3, select F3 from the HOME screen followed by 2 to integrate. This should give you the integral symbol along with the left parenthesis. You should complete this line so that you see the integral symbol followed by (15x^4, x, 1, 3). Entering this command will give you the correct result of 726.
• Use Wolfram Alpha. To integrate 15x4 from x = 1 to 3, enter int 15x^4 from 1 to 3. Wolfram Alpha is very flexible so there are many other ways to enter this integral.
We evaluated the definite integral of (2x + 3)5 from x = 1 to 2 in two ways.
• Using a TI calculator quickly gave us the result of 8502.
• Using the substitution u = 2x + 3 showed that the original definite integral has the same value as the definite integral of (1/2)*u5 from u = 5 to 7. Note that we needed to change the limits of integration when making the substitution. This new definite integral is easier to evaluate. Using the Fundamental Theorem of Calculus (part 2) we obtain (1/12)*76 − (1/12)*56 = 8502.
I solved the following homework problems from section 5.5. We discussed how one thinks about which substitution to make and why some substitutions would fail. Sometimes more than one substitution works.
• (#39) The indefinite integral of sin(2x) / (1 + cos2(x)) dx. We rewrote sin(2x) = 2sin(x)cos(x) and used the substitution u = cos(x). We finished the problem with a second substitution w = 1 + u2. We saw how we could have immediately made the substitution v = 1 + cos2(x), but it is okay to break it up into separate steps.
• (#40) The indefinite integral of sin(x) / (1 + cos2(x)) dx. We used the substitution u = cos(x).
• (#59) The definite integral of e1/x / x2 dx from x = 1 to 2. We used the substitution u = 1/x. Students should try the substitution u = e1/x for homework.
We saw how a limit of Riemann sums reveals how to find the area between curves. If f(x) ≥ g(x) for a ≤ x ≤ b, then the area of the region between these curves for a ≤ x ≤ b is the definite integral from a to b of (f(x) − g(x)) dx. When we ask you to find the area between curves, we usually mean the area of the finite region bounded by the curves. This often involves finding intersection points.

For example I found the exact area of the following region.

• The region bounded by f(x) = x + 3 and g(x) = 5 − x2.
Next we sketched the region bounded by y = 2, y = 0, x = 0, and y = ln(x). For homework determine this area. See if you can do it in two ways – once by integrating with respect to x and once by integrating with respect to y.
Week 13
Apr 9 (Mon) Read sections 4.2 and 6.2. In section 4.2 do #1, 2, 5, 9, 11, 15, 17, 20. Thursday's quiz will cover sections 5.5 and 6.1.

We sketched the region bounded by y = 2, y = 0, x = 0, and y = ln(x). We determined the area of this region in two ways – by integrating with respect to x and then by integrating with respect to y. Our answers were

• Area = 2 + the integral from 1 to e2 of (2 − ln(x)) dx.
• Area = the integral from 0 to 2 of ey dy.
For the first answer, we broke it up into two areas. For 0 ≤ x ≤ 1 we have a rectangle with area 2. For 1 ≤ x ≤ e2 we used a definite integral. At this point we are stuck since we have not yet learned an antiderivative for ln(x). For this reason the second answer is better and gives us a final answer of

Area = e2 − 1.

We solved #12 from section 6.1 which asks for the area of the finite region bounded by the curves 4x + y2 = 12 and x = y. By graphing the functions and finding the intersection points, we noted that we could draw a rectangle of area 72 around the region. Since the area of the region bounded the curves appeared to be about one third of the area of the rectangle, we expected the area of the region to be approximately 72/3 = 24.

We solved this problem exactly in two different ways.

• Integrating with respect to x we see that the area = the definite integral from −6 to 2 of (x − (− sqrt(12 − 4x))) dx + the definite integral from 2 to 3 of (sqrt(12 − 4x) − (− sqrt(12 − 4x))) dx.
• Integrating with respect to y we see that the area = the integral from −6 to 2 of (3 − 0.25y2 − y) dy.
For this problem, integrating with respect to y is much easier to set up and also easier to evaluate the definite integral. We obtain an exact answer of 64/3 = 21.333... which is close to our estimate of 24. I suggest you try to solve other area problems both ways to get a sense of when one approach will be easiest.

We solved #23 from section 6.1 by integrating with respect to x to obtain

• Area = the integral from 0 to π/6 of (cos(x) − sin(2x)) dx + the integral from π/6 to π/2 of (sin(2x) − cos(x)) dx.
We see that it is essential to have a graph here (and on the previous problem) to see the relative position of the curves and to see which intersection points we need.

From section 4.2, the book introduces the Mean Value Theorem along with Rolle's Theorem which is a special case of the Mean Value Theorem. Students should be able to properly state each of these theorems correctly. I drew pictures of functions which were not continuous or not differentiable to see why we need the conditions on continuity and differentiability before stating the conclusion for each theorem. The most important aspect of these theorems is their use in proving other important theorems. In particular I will use the Mean Value Theorem to prove the Fundamental Theorem of Calculus next week.

Using Rolle's Theorem in the following way, we are able to prove that the equation f(x) = x5 + 3x3 + 10x + 10 has exactly one real root (i.e. an x-intercept).

Proof: Let f(x) = x5 + 3x3 + 10x + 10. Since f(−1) is negative, f(0) is positive, and f is continuous everywhere, the Intermediate Value Theorem implies that there is a real root between −1 and 0. What would happen if there were two distinct real roots c1 and c2? Since f is continuous and differentiable everywhere with f(c1) = f(c2) = 0, we would apply Rolle's Theorem to find a value of c between c1 and c2 with f ′(c) = 0. However, this is impossible since f ′(x) = 5x4 + 9x2 + 10 ≥ 10 for all x. Thus there cannot be two (or more) real roots. This shows that there is exactly one real root for f(x) = x5 + 3x3 + 10x + 10.

Apr 11 (Wed) Read section 6.2. In section 6.2 do #2, 6, 7, 9, 12, 14, 16, 17, 33, 55, 56, 58. There will be a quiz Thursday on sections 5.5 and 6.1. There will be two quizzes next week.

We began lecture with the statement, visual interpretation, and finally a proof of both Rolle's Theorem and the Mean Value Theorem. I mentioned that the Mean Value Theorem is often used to prove other theorems. In particular, it can be used to prove the following theorem and its corollary.

• Theorem: If f ′(x) = 0 for all x in the interval (a, b), then f(x) = C (a constant) on the interval (a, b).
• Corollary: If f ′(x) = g ′(x) for all x in the interval (a, b), then f(x) = g(x) + C on the interval (a, b) where C is a constant.
We saw how to calculate volume as V = limn → ∞ Σ A(xk*) Δ x where A(xk*) is the cross-sectional area at xk*. Since the limit of this Riemann sum is by definition a definite integral we obtain
• Volume = the definite integral of cross-sectional area
I graphed the function y = √(x) and shaded in the region R between the x-axis and the curve on the interval [1, 9]. We set up definite integrals for the volume of a solid obtained in each of the following ways.
• Revolving the region R around the x-axis: We get volume = the definite integral from 1 to 9 of π (√(x))2 dx.
• Revolving the region R around the line y = 3: We get volume = the definite integral from 1 to 9 of (π (3)2 − π (3 − √(x))2) dx.
Apr 12 (Thu) Quiz #10 on sections 5.5 and 6.1 will be given during today's discussion section.
Apr 13 (Fri) Read sections 6.3 and 6.5. In section 6.3 do #3, 5, 9, 12, 14, 15, 17, 19, 20. In section 6.5 do #1, 2, 4, 5, 7, 9, 10, 13, 14, 17. Go back and try to solve each area problem (section 6.1) and each volume problem (sections 6.2 and 6.3) in two ways – once by integrating with respect to x and once by integrating with respect to y. There will be two quizzes next week. The quiz on Tuesday will cover sections 4.2, 6.2 and 6.3.

I derived the formula for the average value of a function and showed the geometric interpretation which helped to obtain an approximate value for the average before applying the formula.

We again looked at the region R between the x-axis and the graph of y = √(x) on the interval [1, 9]. We set up definite integrals for the volume of a solid obtained in each of the following ways.

• Letting R form the base of a solid for which cross-sections perpendicular to the x-axis are squares: We get volume = the definite integral from 1 to 9 of (√(x))2 dx.
• Letting R form the base of a solid for which cross-sections perpendicular to the x-axis are equilateral triangles: First we derived the formula A = √(3)/4 * s2 for the area of an equilateral triangle with side length s. We get volume = the definite integral from 1 to 9 of √(3)/4 * (√(x))2 dx.
• Revolving the region R around the y-axis: We get volume = the definite integral from 0 to 1 of (π (9)2 − π (1)2) dy + the definite integral from 1 to 3 of (π (9)2 − π (y2)2) dy.
I used this last example to illustrate another approach called the Cylindrical Shells method. With this approach we obtained
• volume = the definite integral from 1 to 9 of 2π x √(x) dx.
Note that the integrand is simply the cylinder's surface area 2 π r h where for this example the radius r = x and the height h = √(x).

We then found the volumes of two solids using the method of cylindrical shells.

• A region is bounded above by y = 9 − x2 and below by the x-axis on the interval [−3, 3]. This region is revolved around the vertical line x = 5 to form a solid. We get volume = the definite integral from −3 to 3 of 2π (5 − x) (9 − x2) dx.
• A region is bounded above by y = 4x − x2 and below by the x-axis on the interval [1, 4]. This region is revolved around the y-axis to form a solid. We get volume = the definite integral from 1 to 4 of 2π x (4x − x2) dx.
We looked further at the region R between the x-axis and the graph of y = 4x − x2 on the interval [1, 4]. We revolved R around the following lines and determined that integrating with respect to x was a better choice.
• around x = − 2 (do you understand why we used the method of Cylindrical Shells here?)
• around y = 5 (do you understand why we used the method of Washers here?)
Week 14
Apr 16 (Mon) Read sections 3.10 and 4.8. In 3.10 do #6, 23, 24, 25, 26, 28, 31. Quiz #11 is an in-class quiz on sections 4.2, 6.2 and 6.3 which will be given during Tuesday's discussion section. I'll announce information about quiz #12 later this week.

We can find the equation of the tangent line to the graph of f(x) at a particular point. If we call this tangent line L(x), then from the graphs of f(x) and L(x) we see that f(x) ≈ L(x) for x near the point of tangency. We used this approach to approximate the following quantities without a calculator.

• To approximate e0.2, we used the tangent line to the graph of ex at x = 0 to obtain that ex ≈ x + 1 for x near 0. Thus e0.2 ≈ 0.2 + 1 = 1.2. The TI-83 calculator gives the approximation 1.221402758.
• To approximate e−1/10, we used the tangent line to the graph of ex at x = 0 to obtain that ex ≈ x + 1 for x near 0. Thus e−1/10 ≈ −1/10 + 1 = 0.9. The TI-83 calculator gives the approximation 0.904837418.
• To approximate sin(0.05), we used the tangent line to the graph of sin(x) at x = 0 to obtain that sin(x) ≈ x for x near 0. Thus sin(0.05) ≈ 0.05. The TI-83 calculator gives the approximation 0.0499791693. Note that our calculator must be in radian mode.
• To approximate sin(314.2), we used the tangent line to the graph of sin(x) at x = 100π to obtain that sin(x) ≈ x − 100π for x near 100π. Thus sin(314.2) ≈ 314.2 − 100π ≈ 314.2 − 100(3.14159265358979323) = 0.040734641020677. The TI-83 calculator gives the approximation 0.0407233767. Our approximation is good but it was silly to use so many decimal places for π given that our approximation is not correct to that many decimal places. You will learn techniques in MATH 231 (Calculus II) for obtaining a bound on the error term in various approximations.
• To approximate sqrt(4.06), we used the tangent line to the graph of sqrt(x) at x = 4 to obtain that sqrt(x) ≈ 0.25x + 1 for x near 4. Thus sqrt(4.06) ≈ 0.25(4.06) + 1 = 2.015. The TI-83 calculator gives the approximation 2.014944168.
For the examples above we we saw how the graph of the function together with the graph of its tangent line tells us if our approximation will be an underestimate or an overestimate to the exact value of each quantity.

I briefly discussed why it is good to use a polynomial (such as the tangent line) to approximate a given quantity. Polynomials are easier to deal with than some other functions like ex, sin(x), cos(x), etc. When you plug a value into a polynomial, you end up using only addition, subtraction, multiplication and division. This means we can do this more easily by hand. It also means that computers and calculators can more easily perform these calculations. Students who take MATH 231 (Calculus II) will learn more about these polynomials which are referred to as Taylor Polynomials. In this course we will restrict our attention to the use of tangent lines to approximate a function near a point.

I discussed five different methods for approximating the square root of 5.

1. Use the Intermediate Value Theorem to find the positive root of the function f(x) = x2 − 5. Since f(2) = −1 < 0 and f(3) = 4 > 0 we know the root lies between 2 and 3. Now since f(2.2) = −0.16 < 0 and f(2.3) = 0.29 > 0 we know the root lies between 2.2 and 2.3. Now refine further to get the accuracy you desire.
2. Use a linear approximation. Find the tangent line to the graph of f(x) = sqrt(x) at x = 4 to be y = 0.25x + 1. Thus sqrt(x) ≈ 0.25x + 1 for x near 4. We conclude that sqrt(5) ≈ 2.25.
3. Use this algorithm which was often taught to high school students 30 or more years ago.
4. Use Newton's Method to obtain successive estimates for the positive root of the function f(x) = x2 − 5. See section 4.8 for a description of the algorithm used for Newton's Method.
5. Use a calculator.
The third method is mentioned only for historical interest − students will not be tested on this algorithm. We will discuss the fourth method on Wednesday. Students are not allowed the use of a calculator on quizzes, tests or the final exam, but can use one to answer the homework questions from section 4.8. Students should fully understand methods 1, 2 and 4 which apply generally to many different types of functions.
Apr 17 (Tue) Quiz #11 on sections 4.2, 6.2 and 6.3 will be given during today's discussion section.
Apr 18 (Wed) Read section 4.8. In section 4.8 do #11, 12, 13, 15, 18, 19, 29, 31. In section 7.2 do #1–8, 12, 14, 15, 17–31 and 34. When using Newton's Method in the homework from section 4.8, some initial estimates may lead to a sequence of estimates which converge very slowly to a root. Don't be too concerned about obtaining the level of accuracy they seek in the homework. Just make sure that you understand and can apply the iterative process correctly. Many of the integrals in section 7.2 require the use of some basic trigonometric identities along with substitution. You should already have these skills. However I will still discuss these more fully on Friday. Quiz #12 is a take-home quiz on sections 3.10 and 4.8. It will available Friday and due at the beginning of Monday's lecture.

I discussed Newton's Method for finding roots of a function f(x). This is equivalent to finding solutions to f(x) = 0. To solve this you decide upon a suitable first estimate x1. This iterative process then generates successive estimates x2, x3, x4, ... which hopefully converge to one of the roots. For n ≥ 1, the successive estimates are given by

xn + 1 = xn − f(xn) / f ′(xn)

You should know the algorithm as well as the graphical interpretation of how tangent lines are used to generate these successive estimates.

We used Newton's Method to solve the following problems.

• Approximate the square root of 5. We did this by applying Newton's Method to find a root of f(x) = x2 − 5.
• Approximate the x-value for the point of intersection on the graphs of y = 2x3 + x2 and y = x − 1. We did this by applying Newton's Method to find a root of f(x) = 2x3 + x2 − x + 1.
We will see next time what happens graphically and algebraically when your initial estimate is an x-value for which the slope of the curve is 0. If this occurs you should simply begin with a different estimate. We will also see that in some situations Newton's Method may move you further away from a root or cycle around a root without getting any closer with successive estimates.

For homework you will be doing a lot of computations on your calculator. Keep a lot of decimal places in each step until the last step when you finally round off your answer. I showed how to use the calculator in an efficient way so that after an initial set-up, each step in Newton's Method will only take only 3 calculator button pushes. It also keeps all of its decimal places so that you don't have to transcribe so many numbers. To do this on a TI-83 or TI-84, you should do the following set-up.

1. Use the Y = button to enter Y1 = x − f(x) / f ′(x) for your particular function. Thus I entered Y1 = x − (x^2 − 5) / (2x) for our first example.
2. From a blank screen enter Y1. I only see how to do this selecting VARS, Y-VARS, Function, and finally Y1.
3. Finish this line as Y1(initial estimate) using whatever value you wish for your initial estimate. I entered Y1(3) for our first example.
4. Enter Y1(ANS). I had to push the 2nd button on the calculator before accessing ANS which is found by one of your calculator buttons. Note that ANS allows you to use to use the result of a previous computation.
5. Enter ENTRY. I had to push the 2nd button on the calculator before accessing ENTRY which is found by one of your calculator buttons. Note that Entry allows you to access a previous calculator entry without having to retype it.
6. Now repeatedly enter ENTRY to give you successive estimates using Newton's Method. You should keep seeing Y1(ANS) on your calculator for each step in your use of Newton's Method.
You do not have to use the calculator approach listed here. If you want to use this approach but can't follow my instructions as written, then stop by office hours or ask me about it after lecture. It is easy enough to use even if it is difficult to follow here.

We evaluated the integral of each of the six trigonometric functions.

• The integral of sin(x) dx = −cos(x) + C.   [ You should already know this. ]
• The integral of cos(x) dx = sin(x) + C.   [ You should already know this. ]
• The integral of tan(x) dx = ln |sec(x)| + C.   [ Rewrite tan(x) = sin(x) / cos(x) and use the substitution u = cos(x) to obtain the integral of −1/u du = −ln |u| + C = −ln |cos(x)| + C. This answer is acceptable but it can also be rewritten as ln |sec(x)| + C. ]
• The integral of cot(x) dx = −ln |csc(x)| + C.   [ Rewrite cot(x) = cos(x) / sin(x) and use the substitution u = sin(x) to obtain the integral of 1/u du = ln |u| + C = ln |sin(x)| + C. This answer is acceptable but it can also be rewritten as −ln |csc(x)| + C. ]
• The integral of sec(x) dx = ln |sec(x) + tan(x)| + C.   [ Rewrite sec(x) = 1 / cos(x) = cos(x) / cos2(x) = cos(x) / (1 − sin2(x)) and use the substitution u = sin(x) to obtain the integral of 1 / (1 − u2) du. Since 1 / (1 − u2) = 1 / ((1 + u) (1 − u)) = 1/2 (1 / (1 + u) + 1 / (1 − u)), we get 1/2 (ln |1 + u| − ln |1 − u|) + C. Writing this in terms of x and then using a lot of simplification we obtain that the integral of sec(x) dx is ln |sec(x) + tan(x)| + C. Most people just memorize this since it takes so many steps to derive. ]
• The integral of csc(x) dx = ln |csc(x) − cot(x)| + C.   [ An approach similar the one for sec(x) gives this result. ]
We also evaluated the following integrals.
• The integral of sin5(x) cos(x) dx   [ We used the substitution u = sin(x). ]
• The integral of sin5(x) cos3(x) dx   [ We rewrote the integrand as sin5(x) cos2(x) cos(x) = sin5(x) (1 − sin2(x)) cos(x) and then used the substitution u = sin(x). ]
Apr 19 (Thu) Quiz #12 will be a take-home quiz on sections 3.10 and 4.8 due at the beginning of Monday's lecture.
Apr 20 (Fri) Quiz #12 is a take-home quiz on sections 3.10 and 4.8. It should be turned in to Mr. Murphy before the bell rings at the beginning of Monday's lecture.

Use the Test 3 Notes to begin your preparation for Wednesday's test.

We looked back at the integrals for the six trigonometric functions. However for this test you will not be required to know the integral of sec(x) dx or the integral of csc(x) dx.

We discussed strategies for integrating sinn(x) or cosn(x) when n is odd.

• The integral of sin3(x) dx   [ Rewrite the integrand as sin2(x) sin(x) = (1 − cos2(x)) sin(x) and use the substitution u = cos(x). ]
• The integral of sin5(x) dx   [ Rewrite the integrand as sin4(x) sin(x) = (1 − cos2(x))2 sin(x) and use the substitution u = cos(x). ]
• The integral of cos7(x) dx   [ Rewrite the integrand as cos6(x) cos(x) = (1 − sin2(x))3 cos(x) and use the substitution u = sin(x). ]
We discussed strategies for integrating sinn(x) or cosn(x) when n is even.
• The integral of sin2(x) dx   [ Rewrite the integrand using the identity sin2(x) = 1/2 − 1/2 cos(2x) and then use the substitution u = 2x. ]
• The integral of sin4(x) dx   [ Rewrite the integrand as (sin2(x))2 = (1/2 − 1/2 cos(2x))2 = 1/4 − 1/2 cos(2x) + 1/4 cos2(2x). Now use the identity cos2(x) = 1/2 + 1/2 cos(2x) in order to rewrite cos2(2x) as 1/2 + 1/2 cos(4x) in that last term. The rest of the problem can now be solved with some simple substitutions. ]
Next we looked at the following integrals.
• The integral of tan2(x) dx   [ Rewrite the integrand as sec2(x) − 1 and integrate term by term. ]
• The integral of tan4(x) dx = the integral of tan2(x) tan2(x) dx = the integral of tan2(x) (sec2(x) − 1) dx = the integral of (tan2(x) sec2(x) − tan2(x)) dx = the integral of (tan2(x) sec2(x) − (sec2(x) − 1)) dx = the integral of tan2(x) sec2(x) dx − the integral of (sec2(x) − 1) dx. For this first integral use the substitution u = tan(x). The second integral should be clear.
Finally we looked at multiple approaches for the following integrals.
• The integral of tan(x) sec4(x) dx   [ Method 1: Rewrite the integrand as tan(x) sec2(x) sec2(x) = tan(x) (tan2(x) + 1) sec2(x) and use the substitution u = tan(x). Method 2: Rewrite the integrand as sec3(x) sec(x)tan(x) and use the substitution u = sec(x). Method 3: Rewrite the integrand as sin(x) / cos5(x) and use the substitution u = cos(x). Method 4: Use the substitution u = sec2(x).]
• The integral of sin3(x) cos5(x) dx   [ Method 1: Rewrite the integrand as sin2(x) cos5(x) sin(x) = (1 − cos2(x)) cos5(x) sin(x) and use the substitution u = cos(x). Method 2: Rewrite the integrand as sin3(x) cos4(x) cos(x) = sin3(x) (1 − sin2(x))2 cos(x) and use the substitution u = sin(x). ]
Week 15
Apr 23 (Mon)
Apr 25 (Wed) Test 3 (given during lecture)
Apr 26 (Thu) Discussion sections still meet today. For homework read section 3.11 and do #1, 2, 3, 4, 23abcd, 31, 32, 33, 35, 38 from that section.

Section 3.11 includes a lot of details concerning hyperbolic functions. Here are the main details that you need to know for our course. This knowledge should be sufficient for answering the assigned homework questions.

• Definitions for these hyperbolic functions:
• cosh(x) = (ex + e–x) / 2   (Note that cosh(x) and cos(x) are different functions.)
• sinh(x) = (ex – e–x) / 2   (Note that sinh(x) and sin(x) are different functions.)
• tanh(x) = sinh(x) / cosh(x)   (Note that tanh(x) and tan(x) are different functions.)
• Graphs for sinh(x), cosh(x) and tanh(x). The graph of y = cosh(x) is called a catenary. It appears in natural forms such as spider webs. It also appears as a hanging chain.
• Derivatives for these hyperbolic functions:
• ( cosh(x) ) ′ = sinh(x)
• ( sinh(x) ) ′ = cosh(x)
Apr 27 (Fri) Finish the homework from section 3.11. Read the first five pages in section 5.3 on part 1 of the Fundamental Theorem of Calculus. In section 5.3 do #7, 8, 12, 16, 57. Physics and engineering students are encouraged to read section 6.4 and do #1, 2, 7, 8, 9, 13, 15, 18, 19, 20, 21, 24 from that section. Section 6.4 will not be on the final exam.
Week 16 (Last day for U of I classes is Wednesday, May 2, 2012)
Apr 30 (Mon) On the final exam there will be a bonus problem on section 7.1 (Integration by Parts). For practice you may want to look at problems 1 – 36 in section 7.1.

My office hours this week are Tuesday/Thursday 4-5:30pm. The tutoring room will be definitely be open Monday, Tuesday and Wednesday. If there are additional hours I will post them here.

In lecture we used integration by parts to evaluate the following integrals.

• The indefinite integral of x10 ln(x) dx
• The indefinite integral of 6x e2x dx
• The indefinite integral of x2 cos(x) dx
• The indefinite integral of ln(x) dx
• The definite integral of ln(x) dx from 2 to 3
• The definite integral of tan−1(x) dx from 0 to 1
• The indefinite integral of ex sin(x)
May 2 (Wed) The cumulative final exam for section BL1 will be held Thursday, May 10, 1:30 PM – 4:30 PM in 314 Altgeld Hall. Here are the sections covered.
• 50 points of test 1 material (1.1, 1.2, 1.3, 1.5, 1.6, 2.1, 2.2, 2.3, 2.5, 2.6, 2.7, 2.8, trigonometry)
• 50 points of test 2 material (3.1, 3.2, 3.3, 3.4, 3.5, 3.6, 3.7, 3.8, 3.9, 3.11, 4.1, 4.3, 4.4, 4.7)
• 50 points of test 3 material (3.10, 4.2, 4.8, 4.9, 5.1, 5.2, 5.3, 5.4, 5.5, 6.1, 6.2, 6.3, 6.5, 7.2)
• 5 bonus points of material from section 7.1 on integration by parts
There will be no true/false questions. The additional material since the last test which will be included on the final exam are section 3.11 (hyperbolic functions) which will be included with the test 2 material, section 5.3 (Fundamental Theorem of Calculus – part 1) which will be included with the test 3 material, and section 7.1 (integration by parts) which will be included as a bonus question.

See the list of remaining tutoring/office hours for the semester.

Final Exam Period (Friday-Friday, May 4-11, 2012)
May 10 (Thu) Cumulative Final Exam from 1:30 PM – 4:30 PM in 314 Altgeld Hall

 Department of Mathematics College of Liberal Arts and Sciences University of Illinois at Urbana-Champaign 273 Altgeld Hall, MC-382 1409 W. Green Street, Urbana, IL 61801 USA Department Main Office Telephone: (217) 333-3350 Fax (217) 333-9576