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: 2


$\int_{-a}^a f(x) dx = 2\int_0^a f(x) dx$ 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.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.