Calculus: Early Transcendentals 8th Edition

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

Chapter 5 - Section 5.4 - Indefinite Integrals and the Net Change Theorem - 5.4 Exercises - Page 409: 45



Work Step by Step

$$A=\int^{2}_{-1}(x-2|x|)dx$$ From $(-1,0)$, $x\lt0$, therefore $x-2|x|=x-2(-x)=x+2x=3x$. From $[0,2)$, $x\geq0$, therefore, $x-2|x|=x-2x=-x$ So, $A$ would become $$A=\int^{0}_{-1}(3x)dx+\int^{2}_0(-x)dx$$ $$A=(\frac{3x^2}{2})\Bigg]^0_{-1}+(\frac{-x^2}{2})\Bigg]^{2}_0$$ $$A=(0-\frac{3\times(-1)^2}{2})-(\frac{2^2}{2}-0)$$ $$A=-\frac{3}{2}-2$$ $$A=-\frac{7}{2}$$
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