Calculus (3rd Edition)

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



Work Step by Step

We have $$\int_{3}^{0}\left(2t^3-6t^2\right) d t=\frac{2t^{4}}{4}-\frac{6t^3}{3}|_{3}^0\\ =0-\frac{3^4}{2}+2(3)^3=\frac{27}{2}.$$
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