C H A P T E R
The Fundamental Theorem of Calculus
24.1 DEFINITE INTEGRALS AND THE FUNDAMENTAL THEOREM
WeconcludedthepreviouschapterwiththeFundamentalTheoremofCalculus,version1.
If is continuous on [, ], then
is differentiable on , and
.
One reason this result is so exciting is that we can use it to obtain a simple and beautiful method for computing deÞnite integrals. LetÕs look at how this result helps us compute
, where and are constants.
D e f i n i t i o n
Afunction isanantiderivativeof ifitsderivativeis; thatis, isanantideriva-tive of if .
Recallthatiftwofunctionshavethesamederivative,thentheydifferonlybyanadditive constant. In other words, if and are both antiderivatives of (i.e., if , then for some constant . Using this terminology, we can rephrase our last result as follows. Suppose is between and .
761
762 CHAPTER 24 The Fundamental Theorem of Calculus
is an antiderivative of because the derivative of is .
Let be any antiderivative of . Then for some constant .
(Any two antiderivatives of differ only by an additive constant.) It follows that
and .
Suppose that we want to compute . We know that
(by the splitting interval property of deÞnite integrals) Consequently,
(using the endpoint reversal property of deÞnite integrals)
[]
WeÕve shown that .
This is the Fundamental Theorem of Calculus, version 2.
T h e F u n d a m e n t a l T h e o r e m o f C a l c u l u s , v e r s i o n 2
Let be continuous on [, ]. If is an antiderivative of , that is, , then
.
The Fundamental Theorem tells us that to compute the signed area between the graph of and the horizontal axis over the interval [, ] we need only Þnd an antiderivative of
and compute the difference .
Recall that our working deÞnition of is lim 1 , where we partition [, ] into equal pieces, each of length , and label , for 1, , . Calculating is a wonderful alternative to computing
lim 1 !
f
F(b) — F(a)
a b
Figure 24.1
The Fundamental Theorem of Calculus gives us a fantastic amount of power when computing deÞnite integrals. For example, to compute 1 32 , we need to Þnd a function
24.1 DeÞnite Integrals and the Fundamental Theorem 763
whose derivative is 32. The function 3 comes to mind.
3
3 31 1
33 13 26
ItÕs as easy as that! The task of evaluating deÞnite integrals essentially amounts to reversing the process of taking derivatives. ItÕs like the game show ÒJeopardyÓ; our task is to Þnd 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 Þnding the slope of a curve. This is an interesting problem all by itself and certainly worthy of a whole math course. Recently, weÕve been looking at the equally interesting problem of Þnding 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 deÞnite integrals involves the process of antidifferentiatingÑthe process of Þnding derivatives in reverse! When we Þrst started looking at areas, would youhaveguessedthatsuchamarvelousrelationshipwouldexist?Probablynot.Butitdoes. Astounding!
Take a deep breath and think about this for a minute. Then go and explain this amazing result to someoneÑyour roommate, your best friend, your grandmother, your goldÞsh, or all of them!
Using the Fundamental Theorem of Calculus
WeÕll begin by applying the Fundamental Theorem to a familiar example, an example we could do without the Theorem.
EXAMPLE 24.1
SOLUTION
Find the area under 2 5 on the interval 1 to 4.
This is the example we looked at in Section 22.1, Example 22.5 in the following guise: 2 5 is the velocity of a cheetah on the interval [1, 4]. How far has the cheetah traveled from 1 to 4?
The area is 42 5 . To evaluate this using the Fundamental Theorem of Calculus we need an antiderivative of 2 5. The derivative of 2 is 2 and the derivative of 5 is 5, so 2 5 is an antiderivative of 2 5.
4
2 5 41 1
42 5 412 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 ÒJeopardyÓ is in fact harder to play than is the derivative game. In general itÕs simpler to differentiate than it is to Þnd an antiderivative. Try Þnding a function whose derivative is 2 if you doubt this.
764 CHAPTER 24 The Fundamental Theorem of Calculus
v(t)
20 15 10
5 (1, 7)
1 2 3
(4, 13)
4 t
Figure 24.2
In fact, using the Fundamental Theorem of Calculus to solve this problem is the approach that was referred to as the Òsecond mindsetÓ in Chapter 22.1, Example 22.5b. The velocity of the cheetah is given by . Looking for a position function is equivalent to looking for a function whose derivative is . Finding the displacement by computing
change in position, 41, is essentially what we Þnd when computing 41. , and the constants cancel when we compute 41.
EXERCISE 24.1
EXAMPLE 24.2
Any antiderivative of 2 5 can be written in the form 2 5 . Show that when using the Fundamental Theorem to evaluate the integral it doesnÕt matter which antiderivative is used.
NOTATION: The notation is used as a shorthand for . It is convenient because it allows us to write explicitly before evaluating.
Compute 1 2 .
f
f(x) = x2
(1, 1)
1 x
Figure 24.3
SOLUTION We want an antiderivative of 2, i.e., weÕre looking for a function whose derivative is 2. 3 is such a function.
1 2 31 0 0
1 0
1
24.1 DeÞnite Integrals and the Fundamental Theorem 765
EXAMPLE 24.3 Compute 5 7 .
f
e 5 w
Figure 24.4
SOLUTION (the constant factor property), so we can rewrite this.
5 7
5
7
7 ln
We want a function whose derivative is 1 . ln works.
7[ln 5 ln ] 7[ln 5 1]
Noticethatcallingthevariableintheintegrandinsteadof or makesnodifference; the name of the variable has no impact. Also note that the properties of deÞnite 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
!
and
! and !
!
EXAMPLE 24.4
SOLUTION
Suppose that water enters a reservoir at a rate of 40,000 60,000 cos gallons per month, where is measured in months. What is the net change in water level between 0 and 2?
To Þnd 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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