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

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

Question
Applications of integrals
asked 2021-01-15
Calculate the iterated integral.
\(\displaystyle{\int_{{0}}^{{4}}}{\int_{{0}}^{{12}}}{2}{e}^{{{x}+{3}{y}}}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\)

Answers (1)

2021-01-16
\(\displaystyle{I}={\int_{{0}}^{{4}}}{\int_{{0}}^{{12}}}{2}{e}^{{{\left({x}+{3}{y}\right)}}}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\)
Consider the inner integral, name it as \(\displaystyle{I}_{{1}}\)
\(\displaystyle{I}_{{1}}={\int_{{0}}^{{12}}}{2}{e}^{{{x}+{3}{y}}}{\left.{d}{x}\right.}\)
Take the constant 2 out of the integral sign.
\(\displaystyle{I}_{{1}}={2}{\int_{{0}}^{{12}}}{e}^{{{x}+{3}{y}}}{\left.{d}{x}\right.}\)
Step 2
Use substitution,
Substitute t = x + 3y
\(\displaystyle\Rightarrow{\left.{d}{t}\right.}={\left.{d}{x}\right.}\)
Change the limits as per the new variable,
When \(\displaystyle{x}\rightarrow{0},{t}\rightarrow{3}{y}\)
When \(\displaystyle{x}\rightarrow{12},{t}\rightarrow{12}+{3}{y}\)
Rewrite and solve the new integral as shown,
\(\displaystyle{I}_{{1}}={2}{\int_{{{3}{y}}}^{{{12}+{3}{y}}}}{e}^{{t}}{\left.{d}{t}\right.}\)
\(\displaystyle={2}{{\left({e}^{{t}}\right]}_{{{3}{y}}}^{{{12}+{3}{y}}}}\)
\(\displaystyle={2}{\left({e}^{{{12}+{3}{y}}}-{e}^{{{3}{y}}}\right)}\)
Step 3
Use the value of \(\displaystyle{I}_{{1}}\) and solve the outer integral.
\(\displaystyle{I}={\int_{{0}}^{{4}}}{2}{\left({e}^{{{12}+{3}{y}}}-{e}^{{{3}{y}}}\right)}{\left.{d}{y}\right.}\)
Separate the two integrals.
\(\displaystyle{I}={2}{\int_{{0}}^{{4}}}{e}^{{{12}+{3}{y}}}{\left.{d}{y}\right.}-{2}{\int_{{0}}^{{4}}}{e}^{{{3}{y}}}{\left.{d}{y}\right.}\)
\(\displaystyle{I}={2}{e}^{{12}}{\int_{{0}}^{{4}}}{e}^{{{3}{y}}}{\left.{d}{y}\right.}-{2}{\int_{{0}}^{{4}}}{e}^{{{3}{y}}}{\left.{d}{y}\right.}\)
Evaluate the integrals,
\(\displaystyle{I}=\frac{{2}}{{3}}{e}^{{12}}{{\left({e}^{{{3}{y}}}\right]}_{{0}}^{{4}}}-\frac{{2}}{{3}}{{\left({e}^{{{3}{y}}}\right]}_{{0}}^{{4}}}\)
\(\displaystyle{I}=\frac{{2}}{{3}}{e}^{{12}}{\left({e}^{{12}}-{e}^{{0}}\right)}-\frac{{2}}{{3}}{\left({e}^{{12}}-{e}^{{0}}\right)}\)
\(\displaystyle{I}=\frac{{2}}{{3}}{e}^{{12}}{\left({e}^{{12}}-{1}\right)}-\frac{{2}}{{3}}{\left({e}^{{12}}-{1}\right)}\)
Step 4
Combine the terms,
\(\displaystyle{I}=\frac{{2}}{{3}}{\left({e}^{{24}}-{e}^{{12}}-{e}^{{12}}+{1}\right)}\)
\(\displaystyle{I}=\frac{{2}}{{3}}{\left({e}^{{24}}-{2}{e}^{{12}}+{1}\right)}\)
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