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Answers without the blur. Sign up and see all textbooks for free! Q. 46

Expert-verified Found in: Page 1004 ### Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861 # Evaluate each of the double integrals in Exercises 37–54 as iterated integrals. $\int {\int }_{R}{x}^{3}{e}^{{x}^{2}y}dA,\phantom{\rule{0ex}{0ex}}whereR=\left\{\left(x,y\right)|0\le x\le 4and-1\le y\le 1\right\}$

The value is $\frac{{e}^{32}-2{e}^{16}+1}{2{e}^{16}}$

See the step by step solution

## Step 1. Given Information:

Given double integrals :

$\int {\int }_{R}{x}^{3}{e}^{{x}^{2}y}dA,\phantom{\rule{0ex}{0ex}}whereR=\left\{\left(x,y\right)|0\le x\le 4and-1\le y\le 1\right\}$

We want to evaluate each of the double integrals as iterated integrals.

## Step 2. Solution:

Using Fubini's Theorem:

$\int {\int }_{R}{x}^{3}{e}^{{x}^{2}y}dA={\int }_{0}^{4}{\int }_{-1}^{1}{x}^{3}{e}^{{x}^{2}y}dydx\phantom{\rule{0ex}{0ex}}Evaluationprocedurefortheiteratedintegralweget\phantom{\rule{0ex}{0ex}}={\int }_{0}^{4}\left({\int }_{-1}^{1}{x}^{3}{e}^{{x}^{2}y}dy\right)dx\phantom{\rule{0ex}{0ex}}={\int }_{0}^{4}{x}^{3}\left({\int }_{-1}^{1}{e}^{{x}^{2}y}dy\right)dx$

$Firstwesolve\phantom{\rule{0ex}{0ex}}{\int }_{-1}^{1}{e}^{{x}^{2}y}dy\phantom{\rule{0ex}{0ex}}Put{x}^{2}y=tsowehavedy=\frac{dt}{{x}^{2}}\phantom{\rule{0ex}{0ex}}Wheny=1thent={x}^{2}\phantom{\rule{0ex}{0ex}}Wheny=-1thent=-{x}^{2}\phantom{\rule{0ex}{0ex}}Sointegralbecomes:\phantom{\rule{0ex}{0ex}}={\int }_{-{x}^{2}}^{{x}^{2}}\frac{{e}^{t}}{{x}^{2}}dt\phantom{\rule{0ex}{0ex}}=\frac{1}{{x}^{2}}{\int }_{-{x}^{2}}^{{x}^{2}}{e}^{t}dt\phantom{\rule{0ex}{0ex}}=\frac{1}{{x}^{2}}{\left[{e}^{t}\right]}_{-{x}^{2}}^{{x}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{{e}^{{x}^{2}}-{e}^{-{x}^{2}}}{{x}^{2}}\phantom{\rule{0ex}{0ex}}Nowdoubleintegralbecomes:\phantom{\rule{0ex}{0ex}}={\int }_{0}^{4}{x}^{3}\left(\frac{{e}^{{x}^{2}}-{e}^{-{x}^{2}}}{{x}^{2}}\right)dx\phantom{\rule{0ex}{0ex}}={\int }_{0}^{4}x\left({e}^{{x}^{2}}-{e}^{-{x}^{2}}\right)dx\phantom{\rule{0ex}{0ex}}={\int }_{0}^{4}x{e}^{{x}^{2}}dx-{\int }_{0}^{4}x{e}^{-{x}^{2}}dx\phantom{\rule{0ex}{0ex}}={I}_{1}-{I}_{2}$

$Solve{I}_{1}weget\phantom{\rule{0ex}{0ex}}{\int }_{0}^{4}x{e}^{{x}^{2}}dx\phantom{\rule{0ex}{0ex}}Put{x}^{2}=tsowehavexdx=\frac{dt}{2}\phantom{\rule{0ex}{0ex}}whenx=4thent=16\phantom{\rule{0ex}{0ex}}whenx=0thent=0\phantom{\rule{0ex}{0ex}}{I}_{1}becomes{\int }_{0}^{16}\frac{{e}^{t}}{2}dt\phantom{\rule{0ex}{0ex}}={\left[\frac{{e}^{t}}{2}\right]}_{0}^{16}\phantom{\rule{0ex}{0ex}}=\frac{{e}^{16}-1}{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}Solve{I}_{2}weget\phantom{\rule{0ex}{0ex}}Put-{x}^{2}=tsowehavexdx=-\frac{dt}{2}\phantom{\rule{0ex}{0ex}}whenx=4thent=-16\phantom{\rule{0ex}{0ex}}whenx=1thent=0\phantom{\rule{0ex}{0ex}}{I}_{1}becomes{\int }_{0}^{-16}-\frac{{e}^{t}}{2}dt\phantom{\rule{0ex}{0ex}}={\left[-\frac{{e}^{t}}{2}\right]}_{0}^{-16}\phantom{\rule{0ex}{0ex}}=-\frac{{e}^{-16}-1}{2}\phantom{\rule{0ex}{0ex}}=\frac{1-{e}^{-16}}{2}\phantom{\rule{0ex}{0ex}}Sowehave\phantom{\rule{0ex}{0ex}}{I}_{1}-{I}_{2}=\frac{{e}^{16}-1}{2}-\frac{1-{e}^{-16}}{2}\phantom{\rule{0ex}{0ex}}=\frac{{e}^{16}-1}{2}-\frac{{e}^{16}-1}{2{e}^{16}}\phantom{\rule{0ex}{0ex}}=\frac{{e}^{32}-2{e}^{16}+1}{2{e}^{16}}$ ### Want to see more solutions like these? 