# Evaluate the following iterated integrals. int_0^3int_{-2}^1 (2x+3y)dxdy

Evaluate the following iterated integrals.
${\int }_{0}^{3}{\int }_{-2}^{1}\left(2x+3y\right)dxdy$
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

comentezq
$\text{Given integral,}$
${\int }_{0}^{3}{\int }_{-2}^{1}\left(2x+3y\right)dxdy$
$\text{Now}$
${\int }_{0}^{3}{\int }_{-2}^{1}\left(2x+3y\right)dxdy={\int }_{0}^{3}\left(\frac{2{x}^{2}}{2}+3xy{\right)}_{-2}^{1}dy$
$={\int }_{0}^{3}\left({x}^{2}+3xy{\right)}_{-2}^{1}dy$
$={\int }_{0}^{3}\left(\left(1{\right)}^{2}+3\left(1\right)y-\left(-2{\right)}^{2}-3\left(-2\right)y\right)dy$
$={\int }_{0}^{3}\left(1+3y-4+6y\right)dy$
$={\int }_{0}^{3}\left(-3+9y\right)dy$
$=\left(-3y+\frac{9{y}^{2}}{2}{\right)}_{0}^{3}$
$=\left(-3\left(3\right)+\frac{9\left(3{\right)}^{2}}{2}+3\left(0\right)+\frac{9\left(0{\right)}^{2}}{2}\right)$
$=-9+\frac{81}{2}+0$
$=\frac{63}{2}$
$\text{Final answer:}$
${\int }_{0}^{3}{\int }_{-2}^{1}\left(2x+3y\right)dxdy=\frac{63}{2}$