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a+2
By the fundamental theorem,
a+2 a
sin x dx = cos x
since the cosine function has period 2 .
31.3 Compute the mean value V of Theorem 31.3)
x on [0, 4]. For what x in [0, 4] does the value occur (as guaranteed by
4 1 1 4 1/2 1 2 3/2 x x dx = x dx = 4 0 0 4 0 4 3 0 1 3/2 3/2 ) = 1 [( 4)3 0] = 1 (23 ) = 8 = 4 = (4 0 6 6 6 6 3
2 This average value, 4 , is the value of x when x = 4 = 16 . Note that 0 < 16 < 4. 3 3 9 9
CHAP. 31]
THE FUNDAMENTAL THEOREM OF CALCULUS
31.4 Prove the mean-value theorem for integrals (Theorem 31.3).
b 1 f (x) dx b a a Let m and M be the minimum and maximum values of f on [a, b] (The existence of m and M is guaranteed by Theorem 14.2.) Thus, m f (x) M for all x in [a, b], so that Problem 30.3(c) gives
Write
m(b a)
f (x) dx M(b a)
m V M
But then, by the intermediate-value theorem (Theorem 17.4), the value V is assumed by f somewhere in [a, b].
31.5 Prove Theorem 31.1.
Write Then, g(x) g(x + h) g(x) = = =
f (t) dt
x+h a x a
f (t) dt
f (t) dt
a x+h
f (t) dt + f (t) dt
f (t) dt
f (t) dt
[by Theorem 30.4]
x+h x
By the mean-value theorem for integrals, the last integral is equal to hf (x ) for some x between x and x + h. Hence, g(x + h) g(x) = f (x ) h g(x + h) g(x) = lim f (x ) = Dx (g(x)) = lim h h 0 h 0
f (t) dt
Now as h 0, x + h x, and so x x (since x* lies between x and x + h). Since f is continuous,
lim f (x ) = f (x)
and the proof is complete.
31.6 (Change of Variables) Consider a f (x) dx. Let x = g(u), where, as x varies from a to b, u increases or decreases from c to d. [See Fig. 31-1; in effect, we rule out g (u) = 0 in [c, d].] Show that
f (x) dx =
f (g(u))g (u) du
[The right-hand side is obtained by substituting g(u) for x, g (u) du for dx, and changing the limits of integration from a and b to c and d.]
Fig. 31-1
THE FUNDAMENTAL THEOREM OF CALCULUS
[CHAP. 31
Let The chain rule gives
F(x) =
f (x) dx
or F (x) = f (x)
Du (F(g(u))) = F (g(u))g (u) = f (g(u))g (u) Hence, By the fundamental theorem,
f (g(u))g (u) du = F(g(u))
f (g(u))g (u) du = F(g(u))
= F(g(d)) F(g(c))
= F(b) F(a) =
f (x) dx
31.7 Calculate
x 2 + 1 x dx.
Let us nd the antiderivative of
x 2 + 1 x by making the substitution u = x 2 + 1. Then, du = 2x dx, and 1 1 u du = 2 2 u1/2 du = 1 u3/2 2 3
x 2 + 1 x dx =
1 1 1 = u3/2 = (x 2 + 1)3/2 = ( x 2 + 1)3 3 3 3 Hence, by the fundamental theorem,
x 2 + 1 x dx =
1 1 1 ( x 2 + 1)3 = (( 12 + 1)3 ( 02 + 1)3 ) 3 3 0 3 1 3 1 = (( 2) ( 1) ) = (2 2 1) 3 3
algebra
( 2)3 = ( 2)2 2 = 2 2
( 1)3 = 13 = 1
Alternate Method: Make the same substitution as above, but directly in the de nite integral, changing the limits of integration accordingly. When x = 0, u = 02 + 1 = 1; when x = 1, u = 12 + 1 = 2. Thus, the rst line of the computation above yields
1 x 2 + 1 x dx = =
1 3 1 (( 2) ( 1)3 ) = (2 2 1) 3 3
2 3/2 2 1 1/2 du = 1 u u = u3/2 2 1 2 3 3 1 2 1 2
31.8 (a) If f is an even function (Section 7.3), show that, for any a > 0,
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