Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 5: Integrals - Section 5.6 - Definite Integral Substitutions and the Area Between Curves - Exercises 5.6 - Page 303: 11

Answer

$a.\quad \displaystyle \frac{506}{375}$ $b.\quad\displaystyle \frac{86,744}{375}$

Work Step by Step

$a.$ Applying th.7, let $u=g(t)=4+5t$ (a continuous function). $(t=\displaystyle \frac{u-4}{5})$ Then, $\displaystyle \quad du=g'(t)dt=5dt,\qquad dt= \frac{du}{5} $ and, the borders change to $g(0)=4$ and $g(1)=9$. $\displaystyle \int_{0}^{1}t\sqrt{4+5t}dt=\int_{4}^{9} \frac{u-4}{5}\cdot\sqrt{u}\cdot\frac{du}{5} $ $=\displaystyle \frac{1}{25}\int_{4}^{9}\left[u\sqrt{u}-4 \sqrt{u} \right] du$ $=\displaystyle \frac{1}{25}\int_{4}^{9}\left[u^{3/2}-4 u^{-1/2} \right] du$ $=\displaystyle \frac{1}{25}\left[ \frac{u^{3/2+1}}{3/2+1}-4\cdot\frac{u^{1/2+1}}{1/2+1} \right]_{4}^{9}$ $=\displaystyle \frac{1}{25}\left[ \frac{2u^{5/2}}{5}-\frac{8u^{3/2}}{3} \right]_{4}^{9}$ $=\displaystyle \frac{1}{25}\left[ (\frac{2\cdot 9^{5/2}}{5}-\frac{8\cdot 9^{3/2}}{3})-(\frac{2\cdot 4^{5/2}}{5}-\frac{8\cdot 4^{3/2}}{3}) \right]$ $=\displaystyle \frac{1}{25}\left[ (\frac{2\cdot 3^{5}}{5}-\frac{8\cdot 3^{3}}{3})-(\frac{2\cdot 2^{5}}{5}-\frac{8\cdot 2^{3}}{3}) \right]$ $=\displaystyle \frac{1}{25}\left[ (\frac{2\cdot 243}{5}-\frac{8\cdot 27}{3})-(\frac{2\cdot 32}{5}-\frac{8\cdot 8}{3}) \right]$ $=\displaystyle \frac{1}{25}\cdot (\frac{2\cdot 243\cdot 3-8\cdot 27\cdot 5-2\cdot 32\cdot 3+8\cdot 8\cdot 5}{15}) $ $=\displaystyle \frac{506}{375}$ $b.$ Using the same substitution, $u=4+5t,\ du=5dt,$ the borders here change to $g(1)=9$ and $g(49)=9$. $\displaystyle \int_{0}^{1}t\sqrt{4+5t}dt=\int_{9}^{49} \frac{u-4}{5}\cdot\sqrt{u}\cdot\frac{du}{5} $ Following the same steps as above $=\displaystyle \frac{1}{25}\left[ \frac{2u^{5/2}}{5}-\frac{8u^{3/2}}{3} \right]_{9}^{49}$ $=\displaystyle \frac{1}{25}\left[ (\frac{2\cdot 49^{5/2}}{5}-\frac{8\cdot 49^{3/2}}{3})-(\frac{2\cdot 9^{5/2}}{5}-\frac{8\cdot 9^{3/2}}{3}) \right]$ $=\displaystyle \frac{1}{25}\left[ (\frac{2\cdot 7^{5}}{5}-\frac{8\cdot 7^{3}}{3})-(\frac{2\cdot 3^{5}}{5}-\frac{8\cdot 3^{3}}{3}) \right]$ $=\displaystyle \frac{1}{25}\cdot\frac{2\cdot 7^{5}\cdot 3-8\cdot 7^{3}\cdot 5-2\cdot 3^{5}\cdot 3+8\cdot 3^{3}\cdot 5}{15}$ $=\displaystyle \frac{86,744}{375}$
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