## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 5 - The Integral - Chapter Review Exercises - Page 278: 57

#### Answer

$506$

#### Work Step by Step

Let $u=x+4$, then $du=dx$ and when $x:0\to 5$, then $u:4\to 9$. Now, we have We have \begin{align*} \int_{0}^{5} 15x(x+4)^{1/2} dx&=\int_{4}^{9} 15(u-4)(u)^{1/2} du\\ &=\int_{4}^{9} 15u^{3/2}-60u^{1/2} du\\ &=\frac{15}{5/2}u^{5/2}-\frac{60}{3/2}u^{3/2}|_4^9\\ &=6(3^5-2^5)-40(3^3-2^3)\\ &=506. \end{align*}

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