Calculate the iterated integral. \int_{0}^{4}\int_{0}^{12}2e^{x+3y}dx\ dy

Step 1
${\displaystyle I={\int }_0^4{\int }_0^{12}2e^{x+3y}dxdy}$
Consider the inner integral, name it as ${\displaystyle I_1}$
${\displaystyle I_1={\int }_0^{12}2e^{(x+3y)}dx}$
Take the constant 2 out of the integral sign.
${\displaystyle I_1=2{\int }_0^{12}{e}^{(x+3y)}dx}$
Step 2
Use substitution,
Substitute t=x+3y
${\displaystyle \Rightarrow dt=dx}$
Change the limits as per the new variable,
When ${\displaystyle x\to 0,t\to 3y}$
When ${\displaystyle x\to 12,t\to 12+3y}$
Rewrite and solve the new integral as shown,
${\displaystyle I_1=2{\int }_{3y}^{12+3y}{e}^{t}dt}$
${\displaystyle =2{\left(e^t\right]}_{3y}^{12+3y}}$
${\displaystyle =2(e^{12+3y}-e^{3y})}$
Step 3
Use the value of ${\displaystyle I_1}$ and solve the outer integral.
${\displaystyle I=2{\int }_0^4{e}^{12+3y}dy-2{\int }_0^4{e}^{3y}dy}$
${\displaystyle I=2e^{12}{\int }_0^4{e}^{3y}dy-2{\int }_0^4{e}^{3y}dy}$
Evaluate the integrals,
${\displaystyle I=\frac{2}{3}{e}^{12}{\left(e^{3y}\right]}_0^4-\frac{2}{3}{\left(e^{3y}\right]}_0^4}$
${\displaystyle I=\frac{2}{3}{e}^{12}(e^{12}-e^0)-\frac{2}{3}(e^{12}-e^0)}$
${\displaystyle I=\frac{2}{3}{e}^{12}(e^{12}-1)-\frac{2}{3}(e^{12}-1)}$
Step 4
Combine the terms,