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

Calculate the iterated integral.
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Step 1

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