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: 22


$-\displaystyle \frac{2}{3}$

Work Step by Step

Applying th.7, let $u=g(y)=y^{3}+6y^{2}-12y+9 \quad$ (a continuous function). Then, $g'(y)=3y^{2}+12y-12=3(y^{2}+4y-4).$ Then, $\quad \left[\begin{array}{ll} u=y^{3}+6y^{2}-12y+9 & du=g'(y)dy\\ & du=3(y^{2}+4y-4)dy \\ & (y^{2}+4y-4)d=\dfrac{du}{3} \end{array}\right]$, and, the borders change to $g(0)=0+0-0+9=9$ and $\quad g(1)=1+6-12+9=4$ $\displaystyle \int_{0}^{1}(y^{3}+6y^{2}-12y+9 )^{-1/2}(y^{2}+4y-4)dy=\displaystyle \int_{9}^{4}u^{-1/2}(\frac{du}{3})$ $\displaystyle =\frac{1}{3}\int_{9}^{4}u^{-1/2}du$ $=\displaystyle \frac{1}{3}\left[ \frac{u^{-1/2+1}}{-1/2+1} \right]_{9}^{4}$ $=\displaystyle \frac{1}{3}\left[ \frac{u^{1/2}}{1/2} \right]_{9}^{4}$ $=\displaystyle \frac{2}{3}\left[ u^{1/2} \right]_{9}^{4}$ $=\displaystyle \frac{2}{3}(2-3)$ $=-\displaystyle \frac{2}{3}$
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