Answer
$16$
Work Step by Step
\begin{aligned} \int_{-4}^{4}|x| d x&= \int _{-4}^0-xdx+\int _0^4xdx\\
&= \frac{-1}{2}x^{2}\bigg|_{-4}^0+ \frac{1}{2}x^{2}\bigg|_{0}^4\\
&= 8+8\\
&=16 \end{aligned}
University Calculus: Early Transcendentals (3rd Edition)
by Hass, Joel R.; Weir, Maurice D.; Thomas Jr., George B.
Published by
Pearson
ISBN 10:
0321999584
ISBN 13:
978-0-32199-958-0
Chapter 5 - Section 5.4 - The Fundamental Theorem of Calculus - Exercises - Page 320: 27
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