Answer
$$=\frac{27}{2}.$$
Work Step by Step
We have
$$\int_{3}^{0}\left(2t^3-6t^2\right) d t=\frac{2t^{4}}{4}-\frac{6t^3}{3}|_{3}^0\\
=0-\frac{3^4}{2}+2(3)^3=\frac{27}{2}.$$
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 257: 11
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