Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 5 - Section 5.3 - The Fundamental Theorem of Calculus - 5.3 Exercises - Page 407: 53

Answer

$\int_{0}^{\pi}f(x)~dx = 0$

Work Step by Step

We can evaluate the integral: $\int_{0}^{\pi}f(x)~dx$ $=\int_{0}^{\pi/2}f(x)~dx + \int_{\pi/2}^{\pi}f(x)~dx$ $=\int_{0}^{\pi/2}sin~x~dx + \int_{\pi/2}^{\pi}cos~x~dx$ $=sin~x~\vert_{0}^{\pi/2} + cos~x~\vert_{\pi/2}^{\pi}$ $=(sin~\frac{\pi}{2}-sin~0) + (cos~\pi-cos~\frac{\pi}{2})$ $=(1-0) + (-1-0)$ $= 1-1$ $= 0$
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