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.. The Fundam
Trang 1The 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.
If f is continuous on [a, b], thenaxf (t ) dtis differentiable on (a, b) and
d dx
x
a f (t ) dt = f (x)
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
D e f i n i t i o n
A function F is an antiderivative of f if its derivative is f ; that is, F is an
antideriva-tive 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
Trang 2x c
f (t ) dtis an antiderivative of f (x) because the derivative of
x c
f (t ) dtis f (x)
Let F (x) be any antiderivative of f (x) Then F (x) =cxf (t ) dt + C for some constant C (Any two antiderivatives of f differ only by an additive constant.) It follows that
F (b) =
b
c f (t ) dt + C and F (a) =
a
c f (t ) dt + C
Suppose that we want to computeabf (t ) dt We know that
b
a f (t ) dt =
c
a f (t ) dt +
b c
f (t ) dt (by the splitting interval property of definite
integrals) Consequently,
b
a f (t ) dt =
b
c f (t ) dt −
a c
f (t ) dt (using the endpoint reversal property of
definite integrals)
= F (b) + C − [F (a) + C]
= F (b) − F (a) We’ve shown thatabf (t ) dt = F (b) − F (a)
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 f be continuous on [a, b] If F is an antiderivative of f , that is, F = f , then
b
a f (t ) dt = F (b) − F (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)
Recall that our working definition of abf (t ) dt is limn→∞ ni=1f (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=1f (xi)x!
f
F (b) – F(a)
Figure 24.1
The Fundamental Theorem of Calculus gives us a fantastic amount of power when computing definite integrals For example, to compute33t2dt, we need to find a function
Trang 3F (t )whose derivative is f (t) = 3t2 The function F (t) = t3comes to mind.
3 1 3t2dt = F (3) − F (1)
= 33− 13
= 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
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?
SOLUTION The area is14(2t + 5) dt To evaluate this using the Fundamental Theorem of Calculus we
need an antiderivative of 2t + 5 The derivative of t2is 2t and the derivative of 5t is 5, so
F (t ) = t2+ 5t is an antiderivative of 2t + 5
4 1 (2t + 5) dt = F (4) − F (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
x 2
Trang 4v (t)
5 10 15 20
(1, 7)
(4, 13)
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)
EXERCISE 24.1 Any antiderivative of 2t + 5 can be written in the form F (t) = t2+ 5t + C Show that
when using the Fundamental Theorem to evaluate the integral it doesn’t matter which
antiderivative is used
NOTATION:The notation F (x)
b
ais used as a shorthand for F (b) − F (a) It is convenient because it allows us to write F (x) explicitly before evaluating
1 (1, 1)
f (x) = x2
f
x
Figure 24.3
SOLUTION We want an antiderivative of x2, i.e., we’re looking for a function whose derivative is x2
F (x) = x33 is such a function
1 0
x2dx =x
3 3
1 0
3− 0
Trang 5EXAMPLE 24.3 Computee5w7dw.
f
w
e 5
Figure 24.4
SOLUTION abk f (x)dx = kabf (x)dx(the constant factor property), so we can rewrite this
5 e
7
wdw = 7
5 e
1
wdw We want a function whose derivative isw1 F (w) = ln w works
= 7
ln w
5 e
= 7[ln 5 − ln e]
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
b
a k f (x)dx = k
b a
dxk f (x) = k d
dxf (x)
b
a f (x) + g(x) dx =
b
b a
dx(f + g) =df
dx
month, where t is measured in months What is the net change in water level between t = 0 and t = 2?
SOLUTION 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.
Trang 62 0 (40,000 + 60,000 cos t) dt
=
2
0 10,000(4 + 6 cos t) dt
= 10,000
2 0
= 10,000
2
0 4 dt + 6
2 0 cos t dt
additive integrand property
= 10,000
4t
2 0 +6(sin t)
2 0
sin t is an antiderivative of cos t
= 10,0008 − 0 + 6 sin 2 − 6 sin 0
= 10,0008 + 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
1 1+x 2 − 3x9
dx
SOLUTION
1 0
1
1 + x2− 3x9
dx =
1 0
1
1 + x2dx − 3
1 0
x9dx
For the first definite integral we are looking for a function whose derivative is 1
1+x 2 arctan x
is such a function For the second we are looking for a function whose derivative is x9.x1010
is such a function
1 0
1
1 + x2 − 3x9
dx = arctan x
1
0− 3 ·x
10 10
1 0
= [arctan 1 − arctan 0] − 3 1
10
4 − 0 − 3 1
10
10 How did we know that an antiderivative of 1
1+x 2 is arctan x? We had to remember the
derivative of arctan x (It’s actually a bit surprising that every antiderivative of 1
1+x 2 is of the form arctan x + C.)
How did we know that an antiderivative of x9wasx1010? We might make a first guess of
x10but the derivative of x10is 10x9 This is almost what we want, but we don’t want the
Trang 7constant of 10 out front To get rid of it, we divide by 10: dxd 101x10=101 · 10x10= x10,
as desired
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 prob-lems,01x2dx(Example 24.2), and0πsin x dx, without using the Fundamental Theorem
of Calculus The purpose of this exercise is both to encourage you to recall the limit
defini-tion of the definite integral (essential in applicadefini-tions and in instances where we can’t find an
antiderivative) and to help you develop an appreciation for the power of the Fundamental Theorem
Theorem of Calculus
SOLUTION We begin by partitioning [0, 1] into n equal subintervals labeled as shown
∆ x ∆ x =
∆ x
∆ x
x0 x1 x2 x3 x n–1 x n
0
.
1
n
2
n
3
n
n–1
n n
1
n
Figure 24.5
Each subinterval has width x =1n xk= kx =kn, for k = 0, 1, , n, so
x0= 0, x1= 1/n, x2= 2/n, , xn= n/n = 1
1
0 x2dxis defined to be limn→∞Rn We begin by constructing Rn, the general right-hand sum Let f (x) = x2and refer to Figure 24.6
∆ x
x y
x1x2x3 x n
f (x n)
Figure 24.6
The height of the first rectangle is f (x1), the height of the second is f (x2), and so on
1
0 x2dx = limn→∞Rn= limn→∞ ni=1f (xi)x
Trang 81 0
x2dx = lim n→∞
f (x1)x + f (x2)x + f (x3)x + · · · + f (xn)x
= lim n→∞
f (x1) + f (x2) + f (x3) + · · · + f (xn)x
= lim n→∞
(x1)2+ (x2)2+ (x3)2+ · · · + (xn)2
x (because f (x) = x2)
= lim n→∞
1 n
2 + 2 n
2 + 3 n
2 + · · · + n
n
2 1
n (because xk=nk, x =1
n)
= lim n→∞
12+ 22+ 32+ · · · + n2
n3
In order to evaluate this limit we must express 12+ 22+ 32+ · · · + n2in 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
by mathematical induction.3We use it to obtain01x2dx = limn→∞n(n+1)(2n+1)6n3
To evaluate this limit think back to the work we’ve done with rational functions and consider limx→∞ x(x+1)(2x+1)6x3 The degree of the numerator and denominator are equal (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)
6 =1
3. Therefore,
lim n→∞
n(n + 1)(2n + 1)
3.
We conclude that
1 0
x2dx =1
1
sin x
x
one arc of sin x π
Figure 24.7
3 Refer to Appendix D: Proof by Induction.
4 For large n this fraction “looks like”2x3, the dominant term of the numerator over the dominant term of the denominator.
Trang 9SOLUTION Consider 0πsin 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 x3 . x n = π
Each subinterval has width x =πn xk= kx =kπn, for k = 0, , n, so
x0= 0, x1= π/n, x2= 2π/n, x3= 3π/n, , xn= nπ/n = π
Let f (x) = sin x We know
π
0 sin x dx = lim
n→∞Rn= lim
n→∞
n
i=1
f (xi)x
π
0 sin x dx = lim
n→∞
f (x1)x + f (x2)x + f (x3)x + · · · + f (xn)x
= lim n→∞
f (x1) + f (x2) + f (x3) + · · · + f (xn)x
= lim n→∞
sin x1+ sin x2+ sin x3+ · · · + sin xn
x
π
0 sin x dx = lim
n→∞
sin(π/n) + sin(2π/n) + sin(3π/n) + · · · + sin(nπ/n)(π/n)
We look up, but no manna drops down from heaven Alas, we have no way of ex-pressing [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 ap-proximate0πsin x dx There are problems at the end of this section asking you to do this
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
0 sin x dx = lim
n→∞
n
k=1
sin kπ n
π n
If we do use the Fundamental Theorem of Calculus, the problem of computing0πsin x dx becomes straightforward We look for a function whose derivative is sin x We might first guess cos x dxd cos x = − sin x, which is almost what we want dxd (− cos x) = sin x Therefore
π
0 sin x dx = − cos x
π 0
= − cos π − (− cos 0)
= −(−1) + 1
= 2
Trang 10Pause 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
is e−x2 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
0 sin x dx = lim
n→∞
n
k=1
sin kπ n
π
n,
we know that
π
0 sin x dx ≈
n
k=1
sin 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
camel’s displacement from time t = a to time t = b?
SOLUTION 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 a
f (t ) dt
On the other hand, f (t) is the velocity function, so f (t) = dsdt, 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