Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 5 - Section 5.3 - The Fundamental Theorem of Calculus - 5.3 Exercises - Page 400: 51

Answer

$\int_{0}^{\pi} sin~x~dx = 2$

Work Step by Step

We can find the area under the graph: $\int_{0}^{\pi} sin~x~dx$ $=-cos~x~\vert_{0}^{\pi}$ $=-cos(\pi)-[-cos(0)]$ $=-(-1)-(-1)$ $=1+1$ $=2$
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