Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 5 - Integration - 5.4 Working with Integrals - 5.4 Exercises - Page 381: 2

Answer

$\int_{-a}^a f(x) dx = 2\int_0^a f(x) dx$ due to symmetry. Due to symmetry, the integral from $-a$ to $0$ is equal to the integral from $0$ to $a$ because an even function only consists of positive values. Thus, we can simply evaluate the integral from $0$ to $a$ and multiply by 2.

Work Step by Step

Because the function $f$ is even: $\int_{-a}^a f(x) dx = \int_{-a}^0 f(x) dx + \int_0^a f(x) dx = n + n = 2n$, where $n$ is the value obtained from evaluating the integral.
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