# Evaluate the following iterated integrals. int_1^3int_1^2(y^2+y)dx dy

Cem Hayes 2021-01-28 Answered
Evaluate the following iterated integrals.
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$\text{Let the given integral be I -}$
$I={\int }_{1}^{3}\left({y}^{2}+y\right)\left[{\int }_{1}^{2}dx\right]dy$
$\text{Integrating with respect to x,}$
$I={\int }_{1}^{3}\left({y}^{2}+y\right)\left[x{\right]}_{1}^{2}dy$
$I={\int }_{1}^{3}\left({y}^{2}+y\right)\left(2-1\right)dy$
$I={\int }_{1}^{3}\left({y}^{2}+y\right)dy$
$\text{Integrating equation with respect to y,}$
$I={\int }_{1}^{3}\left({y}^{2}+y\right)dy$
$I=\left[\frac{{y}^{3}}{3}+\frac{{y}^{2}}{2}{\right]}_{1}^{3}$
$I=\left(\frac{{3}^{3}}{3}+\frac{{3}^{2}}{2}\right)-\left(\frac{{1}^{3}}{3}+\frac{{1}^{2}}{2}\right)$
$I=\frac{27}{3}+\frac{9}{2}-\frac{1}{3}-\frac{1}{2}$
$I=\frac{54+27-2-3}{6}$
$I=\frac{76}{6}$
$I=\frac{38}{3}$
$\text{Hence,}$
${\int }_{1}^{3}{\int }_{1}^{2}\left({y}^{2}+y\right)dxdy=\frac{38}{3}$