Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 5 - The Integral - 5.2 The Definite Integral - Exercises - Page 245: 39



Work Step by Step

$\int^{1}_{-3}(7t^{2}+t+1)dt=7\int^{1}_{-3}t^{2}dt+\int^{1}_{-3}tdt+\int^{1}_{-3}dt $ $=7\times[\frac{t^{3}}{3}]^{1}_{-3}+[\frac{t^{2}}{2}]^{1}_{-3}+[t]^{1}_{-3}$ $=7(\frac{1^{3}}{3}-\frac{(-3)^{3}}{3})+(\frac{1^{2}}{2}-\frac{(-3)^{2}}{2})+(1-(-3))$ $=7(\frac{1}{3}+\frac{27}{3})+(\frac{1}{2}-\frac{9}{2})+(1+3)$ $=\frac{7\times28}{3}-4+4=\frac{196}{3}$
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