1549901369-Elements_of_Real_Analysis__Denlinger_

410 Chapter 7 • The Riemann Integral

on [a, b], u[a, b] is a closed interval I containing c and d (see Corollary 5.3.12).

Since g is continuous on I, we can define

G(x) = fcx g(u)du, for u EI.

By FTC-II, G is differentiable and G' = g on I. Consider the function

h =Gou. Then h: [a, b] --+IR, and by the chain rule, Vt E (a, b),

h'(x) = G'(u(x))u'(x)

= g(u(x))u'(x).

Since the composition of continuous functions is continuous, h is continuous

on [a, b], and is an antiderivative of g(u(x))u'(x) on (a, b). Therefore, by FTC-I,

J:(g(u(x))u'(x) dx = h(b) - h(a)

= G(u(b)) - G(u(a))

= fcu(b) g - fcu(a) g

ru(b)

= Ju(a) g (by Theorem 7.6.5).

Example 7.6.14 Evaluate {"fox sin(x^2 + ~ )dx.

lo 2

Solution: We let u(x) = x^2 + ~ and g(x) = sinx. Then u'(x) = 2x and

(fa 7r (fa 1

lo xsin(x

2

+ 2 )dx = lo g (u(x))

2

u'(x)dx

= ~ 1fa (go u)(x)u'(x)dx.

Now u(O) = ~ and u( ft) =^3 ;, so we want

u(fa) ~

~ 1 g(u)du = ~ r sin(u)du

u0) li

= ~ [-cos^3 ; +cos~]= ~[0-0] =0.

In elementary calculus, we would probably use a less formal approach. We
would let u = x^2 + ~. Then du = 2xdx and we write simply

J xsin(x^2 + ~)dx = J sinu ·~du= ~[-cosu] + C.