Calculus: Early Transcendentals 8th Edition

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

Chapter 5 - Section 5.2 - The Definite Integral - 5.2 Exercises - Page 390: 47



Work Step by Step

We can write the expression as a single integral: $\int_{-2}^{2}f(x)~dx+\int_{2}^{5}f(x)~dx-\int_{-2}^{-1}f(x)~dx$ $= \int_{-2}^{2}f(x)~dx-\int_{-2}^{-1}f(x)~dx+\int_{2}^{5}f(x)~dx$ $= \int_{-1}^{2}f(x)~dx+\int_{2}^{5}f(x)~dx$ $=\int_{-1}^{5}f(x)~dx$
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