Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 4 - Integrals - 4.5 The Substitution Rule - 4.5 Exercises - Page 346: 46


$$\int_{-\pi/3}^{\pi/3}x^4\sin xdx=0$$

Work Step by Step

Denote the function under the integral $f(x)=x^4\sin x$. Now we have $$f(−x)=(-x)^4\sin(-x)=x^4(-\sin x)=-x^4\sin x=−f(x)$$ i.e. the function $f$ is odd. Knowing that the integral of an odd function with the symmetric integration bounds (one bound is $−a$ and the other is $a$) is equal to $0$ we have that: $$\int_{-\pi/3}^{\pi/3}x^4\sin xdx=0$$
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