Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 5 - The Integral - Chapter Review Exercises - Page 279: 80



Work Step by Step

We have $$\int_{-2}^2f(x)dx=\int_{-2}^1f(x)dx+\int_{1}^2f(x)dx=5+8=13.$$ Now, since $f(x)$ is even, then $$\int_{0}^2f(x)dx=13/2.$$ Now, $$\int_{0}^1f(x)dx=\int_{0}^2f(x)dx-\int_{1}^2f(x)dx=(13/2) -5=\frac{3}{2}.$$
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