# Evaluate the following integral. \int_{0}^{2}\int_{x^{2}}^{2x}xy dy dx

Evaluate the following integral.
${\int }_{0}^{2}{\int }_{{x}^{2}}^{2x}xydydx$
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Step 1:To determine
Evaluate:
${\int }_{0}^{2}{\int }_{{x}^{2}}^{2x}xydydx$
Step 2:Calculation
Consider ${\int }_{0}^{2}{\int }_{{x}^{2}}^{2x}xydydx$
$={\int }_{0}^{2}\frac{x{y}^{2}}{2}{\mid }_{{x}^{2}}^{2x}dx$
$={\int }_{0}^{2}\frac{x}{2}\left({\left(2x\right)}^{2}-{\left({x}^{2}\right)}^{2}\right)dx$
$={\int }_{0}^{2}\frac{x}{2}\left(4{x}^{2}-{x}^{4}\right)dx$
$=\frac{1}{2}{\int }_{0}^{2}\left(4{x}^{3}-{x}^{5}\right)dx$
$=\frac{1}{2}\left(\frac{4{x}^{4}}{4}-\frac{{x}^{6}}{6}\right){\mid }_{0}^{2}$
$=\frac{1}{2}\left({x}^{4}-\frac{{x}^{6}}{6}\right){\mid }_{0}^{2}$
$=\frac{1}{2}\left({2}^{4}-\frac{{2}^{6}}{6}-0+0\right)$
$=\frac{1}{2}\left(16-\frac{64}{6}\right)$
$=\frac{1}{2}\left(\frac{96-64}{6}\right)$
$=\frac{1}{2}\left(\frac{32}{6}\right)$
$=\frac{16}{6}$
$=\frac{8}{3}$
Hence, ${\int }_{0}^{2}{\int }_{{x}^{2}}^{2x}xydydx=\frac{8}{3}$