## Calculus: Early Transcendentals (2nd Edition)

Published by Pearson

# Chapter 7 - Integration Techniques - 7.6 Other Integration Strategies - 7.6 Exercises - Page 555: 40

#### Answer

$L = \frac{1}{27}\,\,\,\left[ {{{22}^{\frac{3}{2}}} - 8} \right]$

#### Work Step by Step

$\begin{gathered} \,find\,\,the\,length\,\,of\,\,the\,\,curve \hfill \\ \hfill \\ Let\,\,y = {x^{\frac{3}{2}}} + 8\,\,on\,\,the\,\,interval\,\,\,\,\left[ {0,2} \right] \hfill \\ \hfill \\ \,\,the\,\,length\,of\,the\,\,curve\,\,y = \,\left( x \right)\,\,is \hfill \\ \hfill \\ {\text{ }}\,L = \int_a^b {\sqrt {1 + {{y'}^2}} } \,dx \hfill \\ \hfill \\ y = {x^{\frac{3}{2}}} + 8\,\,\,\,\,\,then\,\,differentiating,\,\,\,y' = \frac{3}{2}\sqrt x \hfill \\ \hfill \\ L = \int_0^2 {\sqrt {1 + \,{{\left( {\frac{3}{2}\sqrt x } \right)}^2}} dx} \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ = \int_0^2 {\sqrt {4 + 9x} } \,dx \hfill \\ \hfill \\ use\,\,the\,\,formula \hfill \\ \hfill \\ \int_{}^{} {\sqrt {ax + b} \,dx} \,\, = \frac{2}{{3a}}\,{\left( {ax + b} \right)^{\frac{3}{2}}} + C \hfill \\ \hfill \\ then \hfill \\ \hfill \\ L = \int_0^2 {\sqrt {4 + 9x} } dx = \frac{1}{2}\,\,\left[ {\frac{2}{{27}}\,{{\left( {4 + 9x} \right)}^{\frac{3}{2}}}} \right]_0^2 \hfill \\ \hfill \\ evaluate\,\,the\,\,{\text{limits}}\,\,and\,\,simplify \hfill \\ \hfill \\ L = \frac{1}{27}\,\,\,\left[ {{{22}^{\frac{3}{2}}} - 8} \right] \hfill \\ \end{gathered}$

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