Calculus 8th Edition

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

Chapter 4 - Integrals - 4.4 Indefinite Integrals and the Net Change Theorem - 4.4 Exercises - Page 337: 41

Answer

$$ \int_{-1}^{2}(x-2|x|) d x =-3.5 $$

Work Step by Step

$$ \int_{-1}^{2}(x-2|x|) d x $$ Since the graph of y=x is negative for $x<0$, and positive for $x \gt 0$ we must split this integral into two parts to assure that we are getting the absolute value of $x$. So, the given integral can be written as follows: $$\begin{aligned} \int_{-1}^{2}(x-2|x|) d x &=\int_{-1}^{0}[x-2(-x)] d x+\int_{0}^{2}[x-2(x)] d x \\ &=\int_{-1}^{0} 3 x d x+\int_{0}^{2}(-x) d x \\ &=3\left[\frac{1}{2} x^{2}\right]_{-1}^{0}-\left[\frac{1}{2} x^{2}\right]_{0}^{2} \\ &=3\left(0-\frac{1}{2}\right)-(2-0)\\ &=-\frac{7}{2} \\ &=-3.5 \end{aligned}$$
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