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

Question
Applications of integrals
Calculate the iterated integral.
$$\displaystyle{\int_{{0}}^{{4}}}{\int_{{0}}^{{12}}}{2}{e}^{{{x}+{3}{y}}}{\left.{d}{x}\right.}{\left.{d}{y}\right.}$$

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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