Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 5 - Section 5.2 - The Definite Integral - 5.2 Exercises - Page 396: 57

Answer

$\int_{-1}^{5}f(x)~dx$

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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