Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 4 - Integrals - 4.4 Indefinite Integrals and the Net Change Theorem - 4.4 Exercises - Page 336: 21



Work Step by Step

$\int^{0}_{-2}(\frac{1}{2}t^4+\frac{1}{4}t^3-t) dt$ Integrating each separate term: $=[\frac{1}{2\times 5}t^5+\frac{1}{4 \times 4}t^4-\frac{1}{2}t^2]|^0_{-2}$ $=[\frac{1}{10}t^5+\frac{1}{16}t^4-\frac{1}{2}t^2]|^0_{-2}$ $=[\frac{1}{10}(0)^5+\frac{1}{16}(0)^4-\frac{1}{2}(0)^2]- [\frac{1}{10}(-2)^5+\frac{1}{16}(-2)^4-\frac{1}{2}(-2)^2]$ $=0- [\frac{1}{10}(-32)+\frac{1}{16}(16)-\frac{1}{2}(4)]$ $=0- [\frac{-32}{10}+1-2]$ $=\frac{32}{10}+1$ $=\frac{42}{10}$ $=\frac{21}{5}$
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