Calculus: Early Transcendentals 8th Edition

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

Chapter 5 - Review - Exercises - Page 422: 4

Answer

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

Work Step by Step

We can use the definition of the integral in Theorem 4 to find the integral: $\int_{a}^{b}f(x)~dx = \lim\limits_{n \to \infty}\sum_{i=1}^{n}f(x_i)\Delta x$ $\Delta x = \frac{b-a}{n}$ $x_i = a+i~\Delta x$ $\lim\limits_{n \to \infty}\sum_{i=1}^{n}sin(x_i)\Delta x = \int_{0}^{\pi}sin~x~dx$ We can evaluate this integral: $\int_{0}^{\pi}sin~x~dx$ $= -cos~x~\vert_{0}^{\pi}$ $= -cos(\pi)-[-cos(0)]$ $= -(-1)-(-1)$ $= 1+1$ $= 2$
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