Answer
$0$
Work Step by Step
We have
$$
\int_{-x}^x(t^3+t)dt=\frac{1}{4}t^4+\frac{1}{2}t^2|_{-x}^x\\
=\frac{1}{4}x^4+\frac{1}{2}x^2-\frac{1}{4}x^4-\frac{1}{2}x^2\\
=0.
$$
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
Published by
W. H. Freeman
ISBN 10:
1464125260
ISBN 13:
978-1-46412-526-3
Chapter 5 - The Integral - 5.4 The Fundamental Theorem of Calculus, Part I - Exercises - Page 258: 42
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