# Calculate the iterated integral. int_0^4int_0^12 2e^(x+3y)dxdy

Calculate the iterated integral.
${\int }_{0}^{4}{\int }_{0}^{12}2{e}^{x+3y}dxdy$
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

Benedict
$I={\int }_{0}^{4}{\int }_{0}^{12}2{e}^{\left(x+3y\right)}dxdy$
Consider the inner integral, name it as ${I}_{1}$
${I}_{1}={\int }_{0}^{12}2{e}^{x+3y}dx$
Take the constant 2 out of the integral sign.
${I}_{1}=2{\int }_{0}^{12}{e}^{x+3y}dx$
Step 2
Use substitution,
Substitute t = x + 3y
$⇒dt=dx$
Change the limits as per the new variable,
When $x\to 0,t\to 3y$
When $x\to 12,t\to 12+3y$
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.
$I={\int }_{0}^{4}2\left({e}^{12+3y}-{e}^{3y}\right)dy$
Separate the two integrals.
$I=2{\int }_{0}^{4}{e}^{12+3y}dy-2{\int }_{0}^{4}{e}^{3y}dy$
$I=2{e}^{12}{\int }_{0}^{4}{e}^{3y}dy-2{\int }_{0}^{4}{e}^{3y}dy$
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,