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

• x, x², x³, x⁴, ...
• 1/x, 1/x², 1/x³, 1/x⁴, ...
• x^1/2, x^1/3, x^1/4, ...
• sin(x), cos(x), tan(x)
• e^x, ln(x)

We used shifting and other techniques to graph the following functions. We also discussed independent and dependent variables, domain and range, horizontal and vertical intercepts, and the terms increasing and decreasing for each of these examples.

• y = (x + 1)² − 3
• P = (8 − 2t)^1/2
• Q = (5w − 14) / (w − 3)

• 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.

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).

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

• 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?

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.

• 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

• TABLE 4
  x	y
  1	2
  3	4
  5	8
  7	16

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*a^x we obtained the formula y = 3*2^x.

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*a^x 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*a^x we have exponential growth if a > 1 and exponential decay if 0 < a < 1. We looked at the graphs of examples y = 10*2^x 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*a^t to obtain the formula P = 50*2^t/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.

We used logarithms to solve the equation 3 + 4*10^2x = 63.

I introduced inverse functions by first looking at the following tables of values for y = x³ and y = x^1/3.

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) = x^1/3. 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

• log_a(x) = y ⇔ a^y = x (note that ⇔ means if and only if)
• log_a(a^x) = x
• a^log_a(x) = x for x > 0
• log_a(x*y) = log_a(x) + log_a(y)
• log_a(x/y) = log_a(x) − log_a(y)
• log_a(x^r) = r*log_a(x)

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

• ln(x) = y ⇔ e^y = x (note that ⇔ means if and only if)
• ln(e^x) = x for all x
• e^ln(x) = x for all x > 0
• ln(x*y) = ln(x) + ln(y)
• ln(x/y) = ln(x) − ln(y)
• ln(x^r) = r ln(x)
• ln(1) = 0
• ln(e) = 1
• e⁰ = 1

• ln(e³) = 3
• ln(1/e) = −1 since e⁻¹ = 1/e
• ln(sqrt(e)) = 1/2 since e^1/2 = sqrt(e)
• ln(1) = 0 since e⁰ = 1
• ln(e) = 1 since e¹ = e

• ln(x*y) = ln(x) + ln(y) since e^{ln(x) + ln(y)} = e^ln(x)*e^ln(y) = x*y
• ln(x/y) = ln(x) − ln(y) since e^{ln(x) − ln(y)} = e^ln(x) / e^ln(y) = x/y
• ln(x^r) = r ln(x) since e^{r ln(x)} = ( e^ln(x) )^r = x^r

We solved for x in the equation 6 + 2^x = 4 + 2^{x + 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. lim_{x → 3} ( x² ) = 9
2. lim_{x → 4} ( (x² − 2x − 8) / (x − 4) ) = 6
3. lim_{x → 0} ( sin(x) / x ) = 1
4. lim_{x → π/2} ( tan(x) ) (we need to break it up into two one-sided limits)
   • lim_{x → π/2 ⁻} ( tan(x) ) = ∞
   • lim_{x → π/2 ⁺} ( tan(x) ) = −∞

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 −∞.

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 ⇔ 2^y = 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.

• log_a x = ln(x) / ln(a)

We determined the value of the following limits.

1. lim_{x → π/2} ( tan(x) ) (we need to break it up into two one-sided limits)
   • lim_{x → π/2 ⁻} ( tan(x) ) = ∞
   • lim_{x → π/2 ⁺} ( tan(x) ) = −∞
2. lim_{x → 7} ( (sqrt(x + 2) − 3)/(x − 7) ) = 1/6

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 next evaluated the following limits by looking at the graph of y = 3 + 2e^−x.

1. lim_{x → 0} ( 3 + 2e^−x ) = 5
2. lim_{x → ∞} ( 3 + 2e^−x ) = 3
3. lim_{x → −∞} ( 3 + 2e^−x ) = ∞

One tool at our disposal is the following theorem.

The Squeeze Theorem: If f(x) ≤ g(x) ≤ h(x) when x is near a (except possibly at a) and lim_{x → a} f(x) = lim_{x → a} h(x) = L then lim_{x → a} g(x) = L.

We did not have time to cover the following planned material for today's lecture. I will discuss some of it at the beginning of Friday's lecture.

Since −1 ≤ sin(1/x) ≤ 1 implies that −x² ≤ x²sin(1/x) ≤ x², we can use the Squeeze Theorem to show that lim_{x → 0} x² sin(1 / x) = 0.

Appendix D includes the following identities which I will prove in lecture if we have time.

• sin(x + y) = sin(x) cos(y) + cos(x) sin(y)
• cos(x + y) = cos(x) cos(y) − sin(x) sin(y)
• sin(x − y) = sin(x) cos(y) − cos(x) sin(y)
• cos(x − y) = cos(x) cos(y) + sin(x) sin(y)

• 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)

Since −1 ≤ sin(1/x) ≤ 1 implies that −x² ≤ x²sin(1/x) ≤ x², we can use the Squeeze Theorem to show that lim_{x → 0} x² sin(1 / x) = 0.

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)

1. lim_{θ → 0} sin(2θ) / (5θcos(θ)) = 2/5 (see reasons below)
2. lim_{x → 3⁺} ln(x − 3) = −∞
3. lim_{x → 4⁻} (x + 2) / (x − 4) = −∞
4. lim_{w → 1} (w⁴ − 1)/(w − 1) = 4

We define f to be continuous at x = a if lim_{x → 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, if g is continuous at a and f is continuous at g(a), f ∘ g is continuous at a.

We found lim_{x → 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.

We then discussed an approach for finding lim_{x → ∞} f(x) or lim_{x → −∞} 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.

• lim_{x → ∞} (2x² + 1) / (5 + 3x²) = lim_{x → ∞} (2 + 1/x²) / (5/x² + 3) = 2/3
• lim_{x → −∞} (9 + x + 4x²) / (x⁵ + 6) = lim_{x → −∞} (9/x⁵ + 1/x⁴ + 4/x³) / (1 + 6/x⁵) = 0
• lim_{x → ∞} 8x¹⁰ / (2x³ + 11) = lim_{x → ∞} 8x⁷ / (2 + 11/x³) = ∞
• lim_{x → ∞} (15x³ + 1) / sqrt(4x⁶ + 2x + 5) = lim_{x → ∞} (15 + 1/x³) / sqrt(4 + 2/x⁵ + 5/x⁶) = 15/2

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)

We discussed an approach for finding lim_{x → ∞} f(x) or lim_{x → −∞} 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.

• lim_{x → ∞} (5 + 3x)² / (2x² + 6)
• lim_{x → ∞} 8x¹⁰ / (2x³ + 11) = lim_{x → ∞} 8x⁷ / (2 + 11/x³) = ∞
• lim_{x → ∞} (15x³ + 1) / sqrt(4x⁶ + 2x + 5) = lim_{x → ∞} (15 + 1/x³) / sqrt(4 + 2/x⁵ + 5/x⁶) = 15/2

Last time we defined f to be continuous at x = a if lim_{x → 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.

We used the Intermediate Value Theorem to explore the location of the roots (i.e. x-intercepts) for f(x) = x³ − 3x² − x + 5.

I discussed the restricted domains for sin(x), cos(x) and tan(x) so that they are one-to-one and have inverses sin ⁻¹(x), cos ⁻¹(x) and tan ⁻¹(x), respectively. We also looked at the graphs of these inverse functions. 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(2/3))
• cot(cos ⁻¹(x))
• sin(2*tan⁻¹(1/5))

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.

We evaluated the following quantities.

• cot(arcsin(3/5))
• cos(2*arctan(5/12))

To evaluate cos(2*arctan(5/12)) we let θ = arctan(5/12) so that tan(θ) = 5/12. Noting that 0 < θ < π/2, we drew a right triangle with one of its interior angles equal to θ. Since tan(θ) = 5/12 we took the opposite side to have length 5 and the adjacent side to have length 12. Pythagorean's Theorem is used to show the hypotenuse has length 13. Now cos(2*arctan(5/12)) = cos(2*θ) = cos²(θ) − sin²(θ) = (12/13)² − (5/13)² = 119/169.

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 lim_{x → ∞} ( e^−x + 3) = 3 so the graph of y = e^−x + 3 has a horizontal asymptote at y = 3.

We found lim_{x → 2⁻} ( 3 / (x − 2) ) = −∞ so that the graph of f(x) = 3 / (x − 2) has a vertical asymptote at x = 2.

For the function g(x) = (5e^x + 6) / (e^x + 3), we found lim_{x → −∞} (5e^x + 6) / (e^x + 3) = 6/3 = 2. We also found lim_{x → ∞} (5e^x + 6) / (e^x + 3) = lim_{x → ∞} (5 + 6/e^x) / (1 + 3/e^x) = 5/1 = 5 by dividing both the numerator and denominator by e^x. Thus the graph of g(x) has one horizontal asymptote at y = 2 and another one at y = 5.

We discussed how to use limits in order to find the slope of a curve (i.e. the slope of its tangent line) at various points. We obtained the slope of the line tangent to the graph of f(x) = x² at x = 1 and then at x = 2.

I defined the derivative of f(x) as

f ′(x) = lim_{h → 0} (f(x + h) − f(x)) / h

We used the limit definition of a derivative to show that f(x) = x² has derivative f ′(x) = 2x. For homework use the limit definition of a derivative to show that f(x) = x³ has derivative f ′(x) = 3x².

We looked at an example where S(t) = 2^t 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) = lim_{h → 0} ( (2^{6 + h} − 2⁶) / h )

We made a table of values to approximate this limit. The correct value is S ′(6) ≈ 44.3614 mm³/month but you would have to make many entries in your table to feel confident about the number of decimal places shown for your answer. 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.

You may want to look over old tests and quizzes from my previous MATH 220 courses. Use the free tutoring in 447 Altgeld Hall from 4-7pm Mondays through Thursdays. No appointment is necessary. We should have twice as many tutors as usual on Monday and Tuesday since I've asked the Wednesday and Thursday tutors to adjust their schedules for next week.

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 e^x looks like e^x. 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) = lim_{h → 0} (f(x + h) − f(x)) / h
2. f ′(x) = lim_{w → x} (f(w) − f(x)) / (w − x)
3. f ′(a) = lim_{x → a} (f(x) − f(a)) / (x − a)

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. The velocity of the ball = the rate of change of its height = h ′(t) = −32t + 96. We found h ′(1) = 64 feet/second and h ′(4) = −32 feet per second. We noted that the velocity can be negative. 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. 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.

You may want to look over old tests and quizzes from my previous MATH 220 courses. Use the free tutoring in 447 Altgeld Hall from 4-7pm Mondays through Thursdays. No appointment is necessary. We should have twice as many tutors as usual on Monday and Tuesday since I've asked the Wednesday and Thursday tutors to adjust their schedules for next week.

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.

I proved the important theorem which states that if a function is differentiable at a point, then it must also be continuous at that point.

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

• lim_{x → 2} (2x² + 2x) / (−3x² + 3) = −4/3
• lim_{x → 0} (2x² + 2x) / (−3x² + 3) = 0
• lim_{x → ∞} (2x² + 2x) / (−3x² + 3) = −2/3
• lim_{x → −1} (2x² + 2x) / (−3x² + 3) = −1/3
• lim_{x → 1⁻} (2x² + 2x) / (−3x² + 3) = ∞
• lim_{x → 1⁺} (2x² + 2x) / (−3x² + 3) = −∞

We discussed the results of the first test along with strategies for improving one's score on future tests. Students who are interested in dropping our 5 credit hour course and adding our full semester 3 credit hour MATH 115 course "Preparation for Calculus" should see a Math Advisor in 313 Altgeld Hall right away. Students also 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)

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
• ( e^x ) ′ = e^x
• ( a^x ) ′ = a^x 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))

We discussed the results of the first test along with strategies for improving one's score on future tests. Students who are interested in dropping our 5 credit hour course and adding our full semester 3 credit hour MATH 115 course "Preparation for Calculus" should see a Math Advisor in 313 Altgeld Hall right away. Students also have the option of adding a second eight week course.

A line is said to be normal to a curve at a point on that curve if the line is perpendicular to the line tangent to the curve at that point. On the graph of f(x) = x², we found the line tangent to the curve at the point (3,9) and the line normal to the curve at the point (3,9).

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 lim_{θ → 0} sin(θ) / θ = 1. Another useful limit is lim_{θ → 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)

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) = a^x is f ′(x) = lim_{h → 0} ( (a^{x + h} − a^x) / h ) = lim_{h → 0} ( (a^x a^h − a^x) / h ) = a^x lim_{h → 0} ( (a^h − 1) / h ) = a^x f ′ (0). Thus the derivative of a^x is a^x multiplied by the slope of the curve y = a^x at x = 0. This slope turns out to be ln(a) so that ( a^x ) ′ = a^x ln(a). In particular we get ( e^x ) ′ = e^x.

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.

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

Given f(x) = sin(x), if we wanted the 6th derivative we could write f ′′′′′′(x) but this is difficult to read. Instead we could refer to this 6th derivative as f⁽⁶⁾(x). We saw in class that f⁽⁶⁾(x) = −sin(x).

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.

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)

• ( f(x)ⁿ ) ′ = n ( f(x) )ⁿ⁻¹ f ′(x)

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
• w = tan⁵(x)
• v = e^3q
• y = 1 / (x³ + 2)⁵
• y = ( 4sin(θ) + 2cos(θ) )^1/2
• y = cot(x⁴ + 8x)
• y = e^sec(x)
• w = ( sin( e^{t3 + 4t} ) )^1/2
• y = e^{3ln( e2ln(x) )}
• P = ( x² / (x⁵ + 1) )⁴

1. Rewrite y = ln(x) as x = e^y. Obtain dx/dy = e^y. Now dy/dx = 1 / (dx/dy) = 1 / e^y = 1 / x.
2. Note that e^ln(x) = x. Take the derivative with respect to x of both sides of this equation to obtain e^ln(x) * ( ln(x) ) ′ = 1. Now ( ln(x) ) ′ = 1 / e^ln(x) = 1 / x.

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 lectures on September 14–17 are used to show that cos(sin⁻¹(x)) = √(1 − x²).)

• ( tan⁻¹(x) ) ′ = 1 / (1 + x²)
• ( sec⁻¹(x) ) ′ = 1 / (x * √(x² − 1))

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

Example 1

• y = ln( x⁵ e^x3 (x⁴ + 3)⁶ )
• y = ln( x⁵ ) + ln( e^x3 ) + ln( (x⁴ + 3)⁶ )
• y = 5 ln(x) + x³ + 6 ln(x⁴ + 3)
• y ′ = 5/x + 3x² + 6*4x³/(x⁴ + 3)

• 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) )

• y = x^x
• ln(y) = ln( x^x )
• 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 ′ = x^x * ( ln(x) + 1 )

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

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

I solved question #15 in section 3.5.

• e^x/y = x − y
• d/dx (e^x/y) = d/dx (x − y)
• e^x/y*d/dx (x/y) = d/dx(x) − d/dx(y)
• e^x/y*(d/dx(x)*y − x*d/dx(y)) / y² = 1 − dy/dx
• e^x/y*(1*y − x*dy/dx) / y² = 1 − dy/dx
• e^x/y*(y − x*dy/dx) = y² (1 − dy/dx)
• y*e^x/y − x*e^x/y*dy/dx = y² − y²*dy/dx
• y²*dy/dx − x*e^x/y*dy/dx = y² − y*e^x/y
• dy/dx*(y² − x*e^x/y) = y² − y*e^x/y
• dy/dx = (y² − y*e^x/y) / (y² − x*e^x/y)

• y = x^cos(x)
• ln(y) = ln( x^cos(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 ′ = x^cos(x) * ( −sin(x)*ln(x) + cos(x)/x )

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.)

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

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 = 3e^x, y(0) = 8   (Solution: y = 3e^x + 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 = 20e^8x)

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

We found solutions to the following differential equations along with the given initial values.

• 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 is 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 is y = 6x² − 16.)
• dy/dx = 4y, y(0) = 5   (Solution: y = Ce^4x. By plugging in (x, y) = (0, 5) we find that C = 5 so that our solution is y = 5e^4x.)
• dy/dx = 3y, y(5) = 2   (Solution: y = Ce^3x. By plugging in (x, y) = (5, 2) we find that C = 2/e¹⁵ so that our solution is y = (2/e¹⁵)e^3x or y = 2e^3x−15.)

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 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 = Ce^kx. 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.

• 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.)
• dq/dw = 12w, q(1) = 25   (Solution: q = 6w² + C. By plugging in (w, q) = (1, 25) we find that C = 19 so that our solution is q = 6w² + 19.)
• dq/dw = 12q, q(1) = 25   (Solution: q = Ce^12w. By plugging in (w, q) = (1, 25) we find that C = 25/e¹² so that our solution is q = (25/e¹²)e^12w or q = 25e^12w−12.)

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 = 200e^0.03t.

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 = Ce^kt.

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*a^x and y = C*e^kx 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.

I solved the first problem and set up the second problem on this related rates worksheet. Finish the second problem and do the third problem on your own. Solutions will be provided soon.

In lecture today I further discussed strategies for solving related rates problems. I solved problem #28 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 / e^2x. We found f ′(x) = (5 − 10x) / e^2x 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 lim_{x → ∞} f(x) = lim_{x → ∞} ( 5x / e^2x ) = 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.

Students should know terms such as critical numbers, absolute maximum, absolute minimum, local maximum and local minimum as well as the Extreme Value Theorem and the Closed Interval Method.

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

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.

Students should know terms such as concave up, concave down, inflection points as well as the first derivative test and the second derivative test.

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

1. lim_{x → 2} ( (x⁵ − 32) / (4x − 8) ) = 20
2. lim_{x → 0} ( (sqrt(1 + x) − 1 − x/2) / x² ) = −1/8
3. lim_{x → 0} ( (1 − cos(x)) / (x + x²) ) = 0
4. lim_{x → π/2⁺} ( 2 tan(x) / sec²(x) ) = 0
5. lim_{x → ∞} ( ln(x) / (2sqrt(x)) ) = 0
6. lim_{x → ∞} ( e^2x / (x² + 3x + 5) ) = ∞
7. lim_{x → 0⁺} ( x ln(x) ) = 0
8. lim_{x → 0} ( 1 / sin(x) − 1 / x ) = 0
9. lim_{x → ∞} ( (1 + 1 / x)^x ) = e

(slowly)   ln(x), ..., x^1/3, x^1/2, x, x², x³, ..., e^x   (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

lim_{x → ∞} ( (4x¹⁰⁰⁰ + 5x⁵⁰ + 10) / (0.001e^2x) ) = 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

• lim_{x → ∞} ( 1 / x )^x2 = 0
• lim_{x → 0} ( cos(x) / x² ) = ∞

You may want to look over old tests and quizzes from my previous MATH 220 courses. Use the free tutoring in 447 Altgeld Hall with regular hours 4-7pm Mondays through Thursdays. No appointment is necessary. We should have twice as many tutors as usual on Monday and Tuesday since I've asked the Wednesday and Thursday tutors to adjust their schedules for next week.

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 this 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 again discussed l'Hospital's Rule and I gave a proof of a special but very widely applied case. We determined the following limits using l'Hospital's Rule where appropriate and other techniques when l'Hospital's Rule is not applicable. You should think about why forms like 0^∞, ∞¹ and 1 ÷ 0 are not indeterminate.

• lim_{x → 0} ( cos(x) / x² ) = ∞
• lim_{x → ∞} ( 1 / x )^x2 = 0
• lim_{x → 1⁺} ( ln(x¹² − 1) − ln(x⁴ − 1) ) = ln(3)

You may want to look over old tests and quizzes from my previous MATH 220 courses. Use the free tutoring in 447 Altgeld Hall with regular hours 4-7pm Mondays through Thursdays. No appointment is necessary. We should have twice as many tutors as usual on Monday and Tuesday since I've asked the Wednesday and Thursday tutors to adjust their schedules for next week.

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 this 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.

Today was treated like office hours with students asking questions to prepare for Wednesday's test.

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

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.

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

• 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

I derived the formula for 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. Students are expected to know the three formulas above. I do not expect you to know a formula for the sum of higher powers.

We discussed the following sums

• Sum from k = 1 to 100 of 6k
• Sum from k = 1 to n of 10k + 4
• Sum from k = 4 to 60 of k
• Sum from k = 1 to n of 3k/n²
• Sum from k = 1 to n of 5k²/n³

We looked at area as a limit. We then used this limit approach 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.

Problems 65, 69, 73, 74 and 75 from section 4.9 are more difficult than our previous problems concerning position, velocity and acceleration. I solved problems 69 (stone dropped from cliff) and 73 (falling raindrop) in lecture.

For n subintervals of equal width, our notation is R_n, L_n and M_n to represent the right, left and midpoint Riemann sums, respectively.

To approximate the area between the graph of f(x) = x² and the x-axis on the interval [2, 8], we computed the following Riemann sums.

• n = 1 subinterval
  • R₁ = 64 * 6 = 384
  • L₁ = 4 * 6 = 24
  • M₁ = 25 * 6 = 150
• n = 3 subintervals
  • R₃ = 16 * 2 + 36 * 2 + 64 * 2 = 232
  • L₃ = 4 * 2 + 16 * 2 + 36 * 2 = 112
  • M₃ = 9 * 2 + 25 * 2 + 49 * 2 = 166
• n = 12 subintervals (computations not shown)
  • R₁₂ = 183.25
  • L₁₂ = 153.25
  • M₁₂ = 167.875
• n = 60 subintervals (computations not shown)
  • R₆₀ = 171.01
  • L₆₀ = 165.01
  • M₆₀ = 167.995

To obtain the exact area, 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 x_k^* was chosen on each interval [ x_k−1, x_k ] in order to generate f(x_k^*). We then used this limit approach with right Riemann sums to evaluate the area between the x-axis and f(x) = x² on the interval [2, 8].

I gave the definition found on page 372 for the definite integral including the term integrable for when the given limit exists and is the same quantity regardless of the particular choice for each x_k^*.

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

For a function which is continuous on a closed interval, if we use limits to evaluate the definite integral of f(x) from a to b, this theorem tells us that we can choose our x_k^* in any appropriate manner to determine the value of the definite integral. In lecture I will mostly choose to use right 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. I also discussed how to write the definite integral as a limit of left Riemann sums or as a limit of midpoint Riemann sums. Since these limits appear 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 comparison property listed below.

For an integrable function f(x),

• 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).

For the definite integral of f(x) = x² − 9 from −3 to 3, I looked at how to write it as a limit of right Riemann sums. Without showing the steps in the evaluation, I stated that this limit has a value of −36. We looked at the graph of f(x) along with the Riemann sums to see why we should expect a negative answer. I mentioned the somewhat obscure fact that a parabolic arch has area 2/3 * base * height. For our example the area is 2/3 * 6 * 9 = 36, but the definite integral is the negative of this area since f(x) is below the x-axis on the interval [−3, 3].

Some students who have seen calculus before mentioned that we can use antiderivatives to determine the value of a definite integral so we used this approach for the previous example. We found an antiderivative of x² − 9 to be 1/3 * x³ − 9x. Plugging in x = 3 to this antiderivative gives −18. Plugging in x = −3 to this antiderivative gives 18. Subtracting these two values gives −18 − 18 = −36 which is the same answer found in our earlier approaches.

Why does this method with antiderivatives work?. Does it always work? 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 solved 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 we were given 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 following important theorem from section 5.3.

Fundamental Theorem of Calculus (part 2):

If f is continuous on [a, b], then the definite integral of f(x) from a to b = F(b) − F(a) where F is any antiderivative of f on the given interval.

Here are some problems we then solved.

• 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 = ???   (Think carefully about this one!)

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

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.

We discussed all the properties of definite integrals found on pages 379 — 381 along with short-cuts for evaluating definite integrals of even and odd functions from −a to a.

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

Fundamental Theorem of Calculus (part 2): If f is continuous on [a, b] and F is any antiderivative of f on that interval, then the definite integral of f(x) from a to b is equal to F(b) − F(a).

We looked at the definite integral of 6 − 2x from 0 to 7. Since 6 − 2x is continuous on [0, 7] we know that it is integrable on [0, 7]. 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 between the graph of 6 − 2x and the x-axis on [0, 7] to see two triangles. We then compute the area of the triangle above the x-axis minus the area of the triangle below the x-axis.
3. We can use antiderivatives along with the Fundamental Theorem of Calculus (part 2).

1. The limit of Riemann sums is too difficult to compute.
2. The region under the graph of e^x2 on [0, 2] is not a simple shape whose area we know.
3. We cannot find a formula for an antiderivative of e^x2.

We can also use technology to quickly approximate the value of 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.

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).

Fundamental Theorem of Calculus (part 2): If f is continuous on [a, b], then the definite integral of f(x) from a to b = F(b) − F(a) where F is any antiderivative of f on the given interval.

We evaluated the definite integral from 1 to 3 of x⁻² dx.

We then discussed why we could not use the Fundamental Theorem of Calculus for the definite integral from −2 to 1 of x⁻² dx. We saw on Friday that students came up with many different incorrect results here. 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 this example, 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.

We examined 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).

I evaluated the following integrals from section 5.4.

• (#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.

• 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 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 somewhat time-consuming. Rewriting the integrand for the third problem would be much 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 for a definite integral.)
4. The definite integral from −1/3 to 1/3 of tan(x) dx. (We solved this using substitution but we could have obtained the answer of 0 more quickly since tan(x) is an odd function.)

From the chain rule we know that ( f(g(x)) ) ′ = f ′(g(x)) g ′(x). Thus we know that the indefinite integral of f ′(g(x)) g ′(x) dx is equal to f(g(x)) + C.

For a given problem it may be difficult to recognize the integrand as being of the form f ′(g(x)) g ′(x). Thus we use the method of substitution to simplify our approach to solving these problems.

We solved the following problems.

• The definite integral from −2 to 2 of ( 3x² + 10x³ cos⁵(x) / (x⁸ + 1) ) dx [ We first wrote the definite integral as the definite integral of 3x² + the definite integral of the odd function 10x³ cos⁵(x) / (x⁸ + 1). ]
• The indefinite integral of x² cos(x³ + 2) dx [ We made the substitution u = x³ + 2. ]
• The indefinite integral of sin(2x) / (1 + cos²(x)) dx [ We used that sin(2x) = 2sin(x)cos(x) and then made the substitution u = cos(x) followed by another substitution w = 1 + u². For another approach, this could have been done directly with one substitution of v = 1 + cos²(x). ]
• The indefinite integral of x⁵ √(x³ + 4) dx [ We rewrote the integrand as x² x³ √(x³ + 4) and then made the substitution u = x³ + 4. Note that x³ = u − 4. ]
• The definite integral from 0 to 1 of 6x(x² + 2)⁵ dx [ We made the substitution u = x² + 2. Note that we have to change the limits of integration since as x goes from 0 to 1, u goes from 2 to 3. ]

1. g is continuous on [a, b]
2. g is differentiable on (a, b)
3. g ′(x) = f(x)

We then looked at the following examples.

• g(x) = the definite integral from 3 to x of 1/t dt has derivative g ′(x) = 1/x
• g(x) = the definite integral from 2 to x of e^t2 dt has derivative g ′(x) = e^x2

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. Today I only had time to state Rolle's Theorem and show its graphical interpretation.

Rolle's Theorem: Suppose the following conditions hold.

1. f is continuous on [a, b]
2. f is differentiable on (a, b)
3. f(a) = f(b)

I began class by looking at the Fundamental Theorem of Calculus (part 1) and answered #16 from 5.3.

We discussed the statement and the visual interpretation for both Rolle's Theorem and the Mean Value Theorem. I discussed the proof of Rolle's Theorem and will discuss the proof of the Mean Value Theorem on Wednesday. I mentioned that the Mean Value Theorem is often used to prove other theorems. We'll see soon how it is used in the proof of the Fundamental Theorem of Calculus. For now we'll use it 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.

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

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.

• Area = 2 + the integral from 1 to e² of (2 − ln(x)) dx.

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 e^y dy.

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.

We solved #23 from section 6.1 which asks for the area of the finite region bounded the curves y = cos(x), y = sin(2x), x = 0 and x = π/2. We integrated 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 saw how to calculate volume as V = lim_{n → ∞} Σ A(x_k*) Δ x where A(x_k*) is the cross-sectional area at x_k*. Since the limit of this Riemann sum is by definition a definite integral we obtain

• Volume = the definite integral of cross-sectional area

• 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 = 4: We get volume = the definite integral from 1 to 9 of (π (4)² − π (4 − √(x))²) dx.
• 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: Determine the correct definite integral for this volume for homework.

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 used this to find the average value of y = x² on the interval [1, 3].

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 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.

• volume = the definite integral from 1 to 9 of 2π x √(x) dx.

Let R be the finite region bounded by the curves y = 3x and y = x². Revolve R around the following lines and in each case calculate the resulting volume in two ways — once by integrating with respect to x and once by integrating with respect to y. For each problem, one computation of the volume will use the method of Cylindrical Shells and the other computation of the volume will use the method of Washers. After today's lecture you should understand when each method applies.

1. around x = 0 (the y-axis)
2. around x = −2
3. around x = 4
4. around y = 0 (the x-axis)
5. around y = −1
6. around y = 12
7. around x = 5/3 (this one is more difficult and will not appear on the test or quiz)

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 e^0.2, we used the tangent line to the graph of e^x at x = 0 to obtain that e^x ≈ x + 1 for x near 0. Thus e^0.2 ≈ 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 e^x at x = 0 to obtain that e^x ≈ 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.

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 e^x, 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.

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

x_{n + 1} = x_n − f(x_n) / f ′(x_n)

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.

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 manner so that after an initial set-up, each step in Newton's Method only requires a few calculator entries. 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.

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 looked back at the integrals for the six trigonometric functions as well as the integrals for sec²(x), csc²(x), sec(x)tan(x) and csc(x)cot(x). For this test you will not be required to know the integral of sec(x) dx or the integral of csc(x) dx.

We began with 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). ]

• 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 cos⁵(x) dx   [ Rewrite the integrand as cos⁴(x) cos(x) = (1 − sin²(x))² cos(x) and use the substitution u = sin(x). ]

• The integral of cos²(x) dx   [ Rewrite the integrand using the identity cos²(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. ]

• 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). ]

• The integral of cot²(x) dx   [ Rewrite the integrand as csc²(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.

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) = (e^x + e^–x) / 2   (Note that cosh(x) and cos(x) are different functions.)
  • sinh(x) = (e^x – 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)

I discussed test 3 results and plan to discuss details about course grades on Monday.

Bring a number 2 pencil to Monday's lecture for course/instructor evaluations.

I briefly went over hyperbolic functions. See what I have listed above on Thursday for what you'll need to know for this course.

We saw that integration by parts comes from the product rule for derivatives. We used integration by parts to evaluate the following integrals.

• The indefinite integral of x¹⁰ ln(x) dx
• The indefinite integral of 6x e^2x 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

• 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

• The indefinite integral of 6x sin(2x) dx
• The definite integral of tan^{− 1}(x) dx from 0 to 1
• The indefinite integral of x⁴ e^2x dx