C

H

24

A

P

T

E

R

The Fundamental Theorem of Calculus

24.1 DEFINITE INTEGRALS AND THE FUNDAMENTAL THEOREM

We concluded the previous chapter with the Fundamental Theorem of Calculus, version 1. x If f is continuous on [a, b], then a f (t) dt is differentiable on (a, b) and  x d f (t) dt = f (x). dx a One reason this result is so exciting is that we can use it to obtain a simple and beautiful method for computing definite integrals. Let’s look at how this result helps us compute b a f (t) dt, where a and b are constants.

Definition A function F is an antiderivative of f if its derivative is f ; that is, F is an antiderivative of f if F  = f .

Recall that if two functions have the same derivative, then they differ only by an additive constant. In other words, if F and G are both antiderivatives of f (i.e., if F  = G = f ), then F (x) = G(x) + C for some constant C. Using this terminology, we can rephrase our last result as follows. Suppose c is between a and b. 761

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x

x

f (t) dt is an antiderivative of f (x) because the derivative of c

f (t) dt is f (x). c

x Let F (x) be any antiderivative of f (x). Then F (x) = c f (t) dt + C for some constant C. (Any two antiderivatives of f differ only by an additive constant.) It follows that  b  a F (b) = f (t) dt + C and F (a) = f (t) dt + C. c

c

b

Suppose that we want to compute a f (t) dt. We know that  c  b  b f (t) dt = f (t) dt + f (t) dt (by the splitting interval property of definite a a c integrals) Consequently,  b  a  b f (t) dt = f (t) dt − f (t) dt (using the endpoint reversal property of a c c definite integrals) = F (b) + C − [F (a) + C] = F (b) − F (a) b We’ve shown that a f (t) dt = F (b) − F (a). This is the Fundamental Theorem of Calculus, version 2.

The Fundamental Theorem of Calculus, version 2 Let f be continuous on [a, b]. If F is an antiderivative of f , that is, F  = f , then  b f (t) dt = F (b) − F (a). a

The Fundamental Theorem tells us that to compute the signed area between the graph of f and the horizontal axis over the interval [a, b] we need only find an antiderivative F of f and compute the difference F (b) − F (a). b  Recall that our working definition of a f (t) dt is limn→∞ ni=1 f (xi )x, where we partition [a, b] into n equal pieces, each of length x, and label xi = a + ix, for i = 1, . . . , n. Calculating F (b) − F (a) is a wonderful alternative to computing limn→∞ ni=1 f (xi )x!

f F(b) – F(a) a

b

Figure 24.1

The Fundamental Theorem of Calculus gives us a fantastic amount of power when 3 computing definite integrals. For example, to compute 1 3t 2 dt, we need to find a function

24.1

Definite Integrals and the Fundamental Theorem

763

F (t) whose derivative is f (t) = 3t 2. The function F (t) = t 3 comes to mind. 

3

3t 2 dt = F (3) − F (1)

1

= 33 − 1 3 = 26 It’s as easy as that! The task of evaluating definite integrals essentially amounts to reversing the process of taking derivatives. It’s like the game show “Jeopardy”; our task is to find a function whose derivative is what we have been given.1 The Fundamental Theorem of Calculus is a truly amazing result. Stop and think about it for a minute. Back in Chapter 5, we set to work on the problem of finding the slope of a curve. This is an interesting problem all by itself and certainly worthy of a whole math course. Recently, we’ve been looking at the equally interesting problem of finding the area under a curve. This seems like a question worthy of another whole separate math course. But we have just found that these two questions are intimately related. The process of evaluating definite integrals involves the process of antidifferentiating—the process of finding derivatives in reverse! When we first started looking at areas, would you have guessed that such a marvelous relationship would exist? Probably not. But it does. Astounding! Take a deep breath and think about this for a minute. Then go and explain this amazing result to someone—your roommate, your best friend, your grandmother, your goldfish, or all of them!

Using the Fundamental Theorem of Calculus We’ll begin by applying the Fundamental Theorem to a familiar example, an example we could do without the Theorem.

N EXAMPLE 24.1 SOLUTION

Find the area under v(t) = 2t + 5 on the interval t = 1 to t = 4. This is the example we looked at in Section 22.1, Example 22.5 in the following guise: v(t) = 2t + 5 is the velocity of a cheetah on the interval [1, 4]. How far has the cheetah traveled from t = 1 to t = 4? 4 The area is 1 (2t + 5) dt. To evaluate this using the Fundamental Theorem of Calculus we need an antiderivative of 2t + 5. The derivative of t 2 is 2t and the derivative of 5t is 5, so F (t) = t 2 + 5t is an antiderivative of 2t + 5.  4 (2t + 5) dt = F (4) − F (1) 1

= (42 + 5 · 4) − (12 + 5 · 1) = 30 This is the same answer we obtained by calculating the area of the trapezoid directly.

1 It turns out that this game of “Jeopardy” is in fact harder to play than is the derivative game. In general it’s simpler to 2 differentiate than it is to find an antiderivative. Try finding a function whose derivative is ex if you doubt this.

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v(t) 20 15 10 5

(4, 13) (1, 7) 1

2

3

t

4

Figure 24.2

In fact, using the Fundamental Theorem of Calculus to solve this problem is the approach that was referred to as the “second mindset” in Chapter 22.1, Example 22.5b. The velocity of the cheetah is given by v(t). Looking for a position function s(t) is equivalent to looking for a function F whose derivative is v(t). Finding the displacement by computing change in position, s(4) − s(1), is essentially what we find when computing F (4) − F (1). F (t) = s(t) + C, and the constants cancel when we compute F (4) − F (1). N EXERCISE 24.1

Any antiderivative of 2t + 5 can be written in the form F (t) = t 2 + 5t + C. Show that when using the Fundamental Theorem to evaluate the integral it doesn’t matter which antiderivative is used. b  NOTATION: The notation F (x) is used as a shorthand for F (b) − F (a). It is convenient a because it allows us to write F (x) explicitly before evaluating.

N EXAMPLE 24.2

Compute

1 0

x 2dx. f f(x) = x2 (1, 1) 1

x

Figure 24.3

SOLUTION

We want an antiderivative of x 2, i.e., we’re looking for a function whose derivative is x 2. 3 F (x) = x3 is such a function. 

1 0

 x 3 1 x dx =  3 0 2

1 −0 3 1 = N 3

=

24.1

N EXAMPLE 24.3

Compute

Definite Integrals and the Fundamental Theorem

765

5

7 e w dw.

f

e

w

5

Figure 24.4

SOLUTION

b a



kf (x)dx = k

5 e

7 dw = 7 w



b a

f (x)dx (the constant factor property), so we can rewrite this.

5

1 dw We want a function whose derivative is w e  5  = 7 ln w 

1 w.

F (w) = ln w works.

e

= 7[ln 5 − ln e]

N

= 7[ln 5 − 1]

Notice that calling the variable in the integrand w instead of t or x makes no difference; the name of the variable has no impact. Also note that the properties of definite integrals we found in Section 22.4 are coming in very handy here and that these properties agree with the properties of derivatives. The following are of particular computational importance.2  

SOLUTION



 f (x) + g(x) dx = a

b

kf (x)dx = k

a

b

a

N EXAMPLE 24.4

b

b

f (x) dx and

 f (x)dx +

a b

g(x) dx and a

d d kf (x) = k f (x) dx dx d df dg (f + g) = + dx dx dx

Suppose that water enters a reservoir at a rate of r(t) = 40,000 + 60,000 cos t gallons per month, where t is measured in months. What is the net change in water level between t = 0 and t = 2? To find the net change we calculate the signed area under the rate of change function.

2 When

applicable, symmetry considerations can save us a lot of work.

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2

(40,000 + 60,000 cos t) dt

0



2

=

10,000(4 + 6 cos t) dt

0



2

= 10,000

(4 + 6 cos t) dt

0



2

= 10,000 





2

4 dt + 6

0

constant factor property

cos t dt

additive integrand property

0

2 2    = 10,000 4t  +6(sin t)  0

sin t is an antiderivative of cos t

0

 = 10,000 8 − 0 + 6 sin 2 − 6 sin 0

 = 10,000 8 + 6 sin 2 Observe that r(t) is the rate function for water entering the reservoir. The function giving the amount of water in the reservoir at time t is an antiderivative of r(t). The net change is the difference between the final and initial amount. While the antiderivative that we use is not necessarily the amount function, it differs from the amount function by a constant, so the net change in amount is preserved. N

N EXAMPLE 24.5 SOLUTION

Compute

1 0

1 1+x 2

 0

− 3x 9 dx. 1



  1  1 1 1 9 dx = − 3x dx − 3 x 9dx 2 1 + x2 0 1+x 0

For the first definite integral we are looking for a function whose derivative is

1 . arctan x 1+x 2 10 derivative is x 9. x10

is such a function. For the second we are looking for a function whose is such a function.    1 1 1 x 10 1  9 − 3x − 3 · dx = arctan x  0 1 + x2 10 0 0   1 0 − = [arctan 1 − arctan 0] − 3 10 10   1 π = −0−3 4 10 =

π 3 − 4 10

How did we know that an antiderivative of

1 1+x 2

is arctan x? We had to remember the

1 derivative of arctan x. (It’s actually a bit surprising that every antiderivative of 1+x 2 is of the form arctan x + C.) 10 How did we know that an antiderivative of x 9 was x10 ? We might make a first guess of x 10 but the derivative of x 10 is 10x 9. This is almost what we want, but we don’t want the

24.1

Definite Integrals and the Fundamental Theorem d dx

constant of 10 out front. To get rid of it, we divide by 10: as desired.

N



1 10 10 x

=

1 10

767

· 10x 10 = x 10,

Appreciating the Fundamental Theorem Before arriving at the Fundamental Theorem of Calculus we did not have a convenient, widely applicable method of evaluating definite integrals. In this section we tackle two probπ 1 lems, 0 x 2dx (Example 24.2), and 0 sin x dx, without using the Fundamental Theorem of Calculus. The purpose of this exercise is both to encourage you to recall the limit definition of the definite integral (essential in applications and in instances where we can’t find an antiderivative) and to help you develop an appreciation for the power of the Fundamental Theorem.

We begin by partitioning [0, 1] into n equal subintervals labeled as shown.

Δx Δx = 1n

x1

x2

x3 . . .

=

0

1 n

2 n

3 n

n –1 n

xn –1 xn

=

x0

=

Δx =

Δx =

SOLUTION

1 Evaluate 0 x 2dx using the definition of the definite integral but not using the Fundamental Theorem of Calculus.

=

N EXAMPLE 24.6

n n =1

Figure 24.5

Each subinterval has width x = n1 . xk = kx = nk , for k = 0, 1, . . . , n, so x0 = 0,

x1 = 1/n,

x2 = 2/n, . . . ,

xn = n/n = 1.

1

2 0 x dx is defined to be lim n→∞ Rn . We begin by constructing Rn , the general right-hand sum. Let f (x) = x 2 and refer to Figure 24.6.

Δx

y

f(xn)

x1 x2 x3

xn

x

Figure 24.6

1 0

The height of the first rectangle is f (x1), the height of the second is f (x2), and so on. n  x 2dx = limn→∞ Rn = limn→∞ i=1 f (xi )x .

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1

 x 2dx = lim f (x1)x + f (x2)x + f (x3)x + · · · + f (xn)x n→∞

0

 = lim f (x1) + f (x2) + f (x3) + · · · + f (xn) x n→∞   (because f (x) = x 2) = lim (x1)2 + (x2)2 + (x3)2 + · · · + (xn)2 x n→∞    2  2  2 2 3 n 1 2 1 1 k + + +···+ (because xk = , x = ) = lim n→∞ n n n n n n n  12 + 22 + 32 + · · · + n2 = lim n→∞ n3

In order to evaluate this limit we must express 12 + 22 + 32 + · · · + n2 in closed form. Although it is not a geometric sum, it can be expressed in closed form as follows. 12 + 22 + 32 + · · · + n2 =

n(n + 1)(2n + 1) 6

This identity has been delivered to you like manna dropped from heaven. It can be proven 1 . by mathematical induction.3 We use it to obtain 0 x 2dx = limn→∞ n(n+1)(2n+1) 6n3 To evaluate this limit think back to the work we’ve done with rational functions and . The degree of the numerator and denominator are equal consider limx→∞ x(x+1)(2x+1) 6x 3 (both are of degree 3), so the limit is given by the fraction formed by the leading coefficients of the numerator and denominator.4 lim

x→∞

x(x + 1)(2x + 1) 2 1 = = . 6x 3 6 3

Therefore, lim

n→∞

We conclude that

n(n + 1)(2n + 1) 1 = . 6n3 3



1 0

N EXAMPLE 24.7

1 x 2dx = . 3

N

Find the area under one arc of sin x without using the Fundamental Theorem of Calculus. sin x 1

one arc of sin x

π

x

Figure 24.7 3 Refer 4 For

to Appendix D: Proof by Induction. 3 large n this fraction “looks like” 2x , the dominant term of the numerator over the dominant term of the denominator. 6x 3

24.1

SOLUTION

Definite Integrals and the Fundamental Theorem

769

π Consider 0 sin x dx. To use the limit definition of the definite integral we begin by chopping [0, π] into n equal subintervals labeled as shown.

0 = x0

x1

x2

xn = π

x3 . . .

Each subinterval has width x = πn . xk = kx = x0 = 0,

x1 = π/n,

x2 = 2π/n,

kπ n ,

for k = 0, . . . , n, so

x3 = 3π/n, . . . ,

xn = nπ/n = π.

Let f (x) = sin x. We know 

π

⎛ ⎞ n  sin x dx = lim Rn = lim ⎝ f (xi )x ⎠ . n→∞

0



π

0

i=1

 sin x dx = lim f (x1)x + f (x2)x + f (x3)x + · · · + f (xn)x n→∞

0



n→∞

 = lim f (x1) + f (x2) + f (x3) + · · · + f (xn) x n→∞ 

= lim sin x1 + sin x2 + sin x3 + · · · + sin xn x n→∞

π

 sin x dx = lim sin(π/n) + sin(2π/n) + sin(3π/n) + · · · + sin(nπ/n) (π/n) n→∞

We look up, but no manna drops down from heaven. Alas, we have no way of expressing [sin(π/n) + sin(2π/n) + sin(3π/n) + · · · + sin(nπ/n)] in closed form; we have a problem evaluating this limit. Without using the Fundamental Theorem, we can only apπ proximate 0 sin x dx. There are problems at the end of this section asking you to do this.

N

EXERCISE 24.2

Although we can write what we’ve done in Example 24.7 using summation notation, that is just shorthand, not a closed form; it is nice and compact, but it does not help us evaluate the limit. As an exercise, express what we’ve done using summation notation and compare your answer with what is given below.    π n  kπ π sin x dx = lim sin n→∞ n n 0 k=1

π If we do use the Fundamental Theorem of Calculus, the problem of computing 0 sin x dx becomes straightforward. We look for a function whose derivative is sin x. We might first d d cos x = − sin x, which is almost what we want. dx (− cos x) = sin x. guess cos x. dx Therefore  π π  sin x dx = − cos x  0

0

= − cos π − (− cos 0) = −(−1) + 1 = 2.

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Pause for a moment to admire what we’ve discovered. The area under one arc of sin x is 2. Exactly. Remarkable! Is the moral of Example 24.7 that the limit definition of the definite integral is merely academic? No! In practice, we use the Fundamental Theorem of Calculus to compute definite integrals exactly whenever we can, but sometimes it is difficult (or impossible) to find an antiderivative of the integrand. (For instance, try to find a function whose derivative 2 is e−x . But don’t try for too long; it’s impossible.5) In this case, the best we can do is to approximate the definite integral. The limit definition of derivative gives us a means of approximation. For example, since    π n  kπ π , sin x dx = lim sin n→∞ n n 0 k=1

we know that  0

π

sin x dx ≈

n 

 sin

k=1

kπ n



π , n

for n large.

While this latter sum is tedious to compute by hand for n large, a computer or programmable calculator can deal with it easily. We’ll return to numerical approximations in Chapter 26. In addition to providing a computational tool for approximating definite integrals, the limit definition allows us to apply the work we’ve done with definite integrals to new situations. This will be our focus in Chapter 27 and again in Chapter 29.

The Fundamental Theorem of Calculus, Take Two The Fundamental Theorem of Calculus tells us that we can find the area under the graph of a continuous function f on the interval [a, b] by finding any antiderivative F of f and evaluating the difference F (b) − F (a). We’ve seen that this is an amazing and powerful result. We now show you (through examples) that if the integrand is thought of as a rate function, then the Fundamental Theorem of Calculus tells us something we’ve known for some time.

N EXAMPLE 24.8 SOLUTION

A camel is walking along a straight desert road with velocity f (t). How can we express the camel’s displacement from time t = a to time t = b? On the one hand we can partition the time interval [a, b] into n equal pieces, approximate the camel’s displacement on each subinterval by assuming constant velocity on the subinterval, sum the displacements, and take the limit as n grows without bound. The camel’s net change in position is given by  b f (t) dt. a

On the other hand, f (t) is the velocity function, so f (t) = ds dt , where s(t) gives the camel’s position at time t. s (t) = f (t), so s(t) is an antiderivative of f (t). We know that if we can find s(t), then the camel’s net change in position must be given by s(b) − s(a). 5 It

is impossible to find such a function unless you’re willing to use an infinite polynomial.

24.1

So

b a

Definite Integrals and the Fundamental Theorem

771

f (t) dt = s(b) − s(a), where s is an antiderivative of f .       net change in position position = − position on [a, b] at t = b at t = a

When we interpret the integrand as a rate function the Fundamental Theorem of Calculus becomes transparent. Because any two antiderivatives of f differ only by an additive constant and the constants will cancel in the course of subtraction, it follows that s(t) could be any antiderivative of f .  b b ds  dt = s(t) = s(b) − s(a) N a a dt

N EXAMPLE 24.9

If f (t) is the rate at which water is flowing into or out of a reservoir and we let W (t) be the amount of water in the reservoir at time t, then dW dt = f (t). W (t) is an antiderivative of f (t). The net change in the amount of water in the reservoir over the time interval [a, b] is given by  b  b dW f (t) dt = dt = W (b) − W (a). dt a a Again, when we interpret the integrand as a rate function, the Fundamental Theorem of Calculus is transparent. It is when the integrand, f (t), is not thought of as a rate function that the Fundamental Theorem is most surprising. N

PROBLEMS FOR SECTION 24.1 1. (a) Using a computer or programmable calculator, find upper and lower bounds for the area under one arc of cos x using Riemann sums. Explain how you can be sure your lower bound is indeed a lower bound and your upper bound is an upper bound. (Do not use the Fundamental Theorem of Calculus to do so.) Your upper and lower bounds should differ by no more than 0.01. (b) Use the Fundamental Theorem of Calculus to show that the area under one arc of the cosine curve is exactly 2. 2. An object’s velocity at time t, t in seconds, is given by v(t) = 10t + 3 meters per second. Find the net distance traveled from time t = 1 to t = 9. Do this in two ways. First, look at the appropriate signed area and solve geometrically, without the Fundamental Theorem. Then calculate the definite integral  9 (10t + 3) dt 1

using the Fundamental Theorem of Calculus. 3. Use the Fundamental Theorem of Calculus to calculate 4. Evaluate

3

1 1 t

dt.

2 1

t 3 dt.

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The Fundamental Theorem of Calculus

5. Find the area under the graph of y = ex between x = 0 and x = 1. 6. Find the area under the graph of y = e−x between (a) x = −1 and x = 0. (Why should the answer be the same as the answer to the previous problem?) (b) x = 0 and x = 1. 7. Aimee and Alexandra spent Friday √ afternoon eating hot cinnamon hearts. If they gobbled hearts at a rate of 1.5t + t hearts per minute, then how many hearts did they consume between time t = 0 and t = 9, t given in minutes? 8. Suppose the temperature of an object is changing at a rate of r(t) = −2e−t degrees Celsius per hour, where t is given in hours. (a) (b) (c) (d)

Is the object heating, or cooling? Between time t = 0 and t = 1, how much has the temperature changed? Between t = 1 and t = 2, how much has the temperature changed? If the object was 100 degrees Celsius at time t = 0, how hot is it at time t = 1?

9. Let f (x) =

x

1 1 t

dt, x > 0. (a) Find f (1), f (5), f (10), and f (1/2).

y=1 t

t

(b) What is an alternative formula for f (x)? x (c) Often mathematicians define the natural logarithm by ln x = 1 1t dt for x > 0. Suppose this was the definition you had been given. Use the Fundamental Theorem of Calculus to show that ln x is increasing and concave down for x > 0. 10. Find the value of x > 1 such that the area under the graph of 1/t from 1 to x is 1.   11. The rate of change of water level in a tank is given by r(t) = 2 sin π4 t gallons per hour, where t is measured in hours. At time t = 0 there are 30 gallons of water in the tank. (a) Between time t = 0 and t = 8, when will the water level in the tank be the highest? (b) What is the maximum amount of water that will ever be in the tank? (c) What is the minimum amount of water that will ever be in the tank? (d) Is the amount of water added to the tank between t = 0 and t = 1 less than, greater than, or equal to the amount lost between t = 4 and t = 5? (Try to answer without doing any computations.)

24.1

12. Let g(x) =

x 0

Definite Integrals and the Fundamental Theorem

773

e−t dt. 2

(a) Where is g(x) zero? Positive? Negative? y y = e –t2 t

(b) Where is g(x) increasing? Decreasing? (c) Where is g(x) concave up? Concave down? (d) Is g(x) even, odd, or neither? (e) Although we cannot find an antiderivative for e−t , we are able to get a lot of information about g(x). Sketch g(x). (f ) Using a computer or programmable calculator, approximate g(1), g(−1), and g(2). Go back and look at all your answers to this question; make sure that they are consistent with one another. 2

13. Let h(x) =

x 0

sin(t 2) dt.

Graph sin(t 2) on the domain [−3, 3]. On (0, ∞), where is h(x) positive? Negative? Is h(x) even, odd, or neither? Explain. On [0, 3] where is h(x) increasing? Decreasing? Give exact answers. What is the absolute maximum value of sin(t 2)? What is the absolute minimum value of sin(t 2)? (f ) Where on [0, 3] does h attain its maximum and minimum values? Will your answers change if the domain is (0, ∞)? If the domain is (−∞, ∞)? (g) Numerically approximate the maximum value of h on [0, 3]. (a) (b) (c) (d) (e)

14. (a) Find

1

−1

|x| dx using the area interpretation of the definite integral.

(b) Show that |x2 | is not an antiderivative of |x| on [−1, 1] by showing that applying the Fundamental Theorem as if it were, gives the wrong answer. (c) Find an antiderivative of |x| on [−1, 0). Find an antiderivative of |x| on (0, 1]. 2

15. Find the following two definite integrals without using the Fundamental Theorem of Calculus. Instead, use the area interpretation of the definite integral. 7 7 (a) −3(π + 1)dx (b) −3 | − 2x − 4|dx 16. Evaluate the following. (If you haven’t done Problem 14, do that first.) 5 3 (b) 0 |(x + 3)(x − 1)|dx (a) −3 |x 2 − 4|dx 17. Calculate the following definite integrals by calculating the limit of Riemann sums. You’ll need to use the formulas provided. Check your answers using the Fundamental Theorem of Calculus.

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CHAPTER 24

The Fundamental Theorem of Calculus



5

(a)

1+2+3+···+n=

xdx 0



5

(b) 

n(n + 1)(2n + 1) 6   n(n + 1) 2 3 3 3 3 1 +2 +3 +···n = 2 12 + 22 + 32 + · · · n2 =

x 2dx

0

(c)

2

n(n + 1) 2

x 3dx

0

18. Read Appendix D: Proof by Induction. Then prove that the formulas provided in the previous problem are valid. 19. Given the graph below, compute are differentiable functions.

b a

[f (x)g(x) + g (x)f (x)]dx. Assume that f and g y 6

(b, 5.5)

f(x)

4 3

(a, 4) a

x

b

–5

(b, –5) g(x)

20. (a) Write a Riemann sum with 10 equal subdivisions that gives an overestimate for the area under ln x on [1, 6]. Write your answer in two ways, once with summation notation and one without summation notation (using + · · · +). State clearly and precisely the meaning of any notation used in your sum. (b) Consider the Riemann sum 49 

ln(3 + k · (5/50)) ·

k=0

5 . 50

This is an underestimate for what integral? (The answer to this question is not unique.) 21. Put the following in ascending order, using < or = as appropriate.  2  2.5  2 ln x dx ln x dx, ln x dx 0.5

1

1

22. Put the following in ascending order, using < or = as appropriate.  6 ln x dx L50 R50 L10 R10 0 1

Explain your reasoning briefly.

24.2

The Average Value of a Function: An Application of the Definite Integral

775

23. (a) Which of the following is an antiderivative of ln x? ii. x ln x iii. x ln x − x iv. arctan(ln x) i. x1 6 (b) Evaluate 1 ln x dx.

24.2 THE AVERAGE VALUE OF A FUNCTION:

AN APPLICATION OF THE DEFINITE INTEGRAL

In the modern world, a data set is often presented in a way to succinctly convey basic information about the central tendency of the data. The average is one measure of central tendency of a data set. We’re familiar with computing averages when given a discrete set of data points. For example, to determine the average score on a quiz, sum all quiz scores and divide by the number of quizzes. In this section we will look at the average value of a continuous function and find that the definite integral enables us to compute this. For example, we will look at how to determine the average velocity of an object given a continuous velocity function and the average temperature of an object given a continuous temperature function.

N EXAMPLE 24.10

A dove flies at a velocity of w(t) = 3t 2 + 4 meters per second on the interval [1, 3]. A hawk 2 flies at a velocity of v(t) = 51 2 (−t + 4t − 3) meters per second over the same interval. (a) How far does each travel during this time? (b) What is the average velocity of each bird on the interval [1, 3]?

SOLUTION

(a) The net change in position (or distance traveled, as velocity is nonnegative here) is given by a definite integral. 

3

Dove:

(3t 2 + 4) dt

1

3  = (t 3 + 4t) 1

= (27 + 12) − (1 + 4)  Hawk: 1

3

= 34 meters    3 51 51 −t 3 2 2 (−t + 4t − 3) dt = + 2t − 3t  2 2 3 1    −1 51 (−9 + 18 − 9) − +2−3 = 2 3   51 −4 = 0− 2 3 = 34 meters