Calculus (3rd Edition)

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

Chapter 5 - The Integral - Chapter Review Exercises - Page 278: 68



Work Step by Step

Let $u=t+8$, then $du=dt$ and $u:9\to16$. Now, we have \begin{align*} \int_1^8 t^2(t+8)^{1/2} dt& =\int_{9}^{16} (u-8)^2u^{1/2} du\\ & =\int_{9}^{16} u^{5/2} -16u^{3/2}+64 u^{1/2}du\\ & =\frac{1}{7/2}u^{7/2}-\frac{16}{5/2}u^{5/2}+\frac{64}{3/2}u^{3/2}|_{9}^{16} \\ & =\frac{2}{7}(16^{7/2}-9^{7/2})-\frac{32}{5}(16^{5/2}-9^{5/2})+\frac{128}{3}(16^{3/2}-9^{3/2})\\ &=\frac{66838}{105}. \end{align*}
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