## Calculus with Applications (10th Edition)

Published by Pearson

# Chapter 9 - Multiveriable Calculus - Chapter Review - Review Exercises - Page 519: 79

#### Answer

$$\frac{1}{2}$$

#### Work Step by Step

\eqalign{ & \int_0^1 {\int_0^{2x} {xydy} dx} \cr & = \int_0^1 {\left[ {\int_0^{2x} {xydy} } \right]dx} \cr & {\text{solve the inner integral by treating }}x{\text{ as a constant}} \cr & = x\int_0^{2x} {ydy} \cr & {\text{integrate }} \cr & = x\left[ {\frac{{{y^2}}}{2}} \right]_0^{2x} \cr & {\text{evaluating the limits in the variable }}y \cr & = x\left[ {\frac{{{{\left( {2x} \right)}^2}}}{2} - \frac{{{{\left( 0 \right)}^2}}}{2}} \right] \cr & {\text{simplifying}} \cr & = \frac{{4{x^3}}}{2} \cr & = 2{x^3} \cr & \cr & \int_0^1 {\left[ {\int_0^{2x} {xydy} } \right]dx} = \int_0^1 {2{x^3}dx} \cr & {\text{integrating}} \cr & = \left[ {\frac{{2{x^4}}}{4}} \right]_0^1 \cr & = \frac{1}{2}\left( {{x^4}} \right)_0^1 \cr & = \frac{1}{2}\left( {1 - 0} \right) \cr & = \frac{1}{2} \cr}

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