91Ó°ÊÓ

Given double integrals :

$$\begin{gathered} ∫{∫}_R{x}^3{e}^{x^{2}y}dA, \\ whereR=\left\{\right(x,y\left)\right|0≤x≤4and−1≤y≤1\} \end{gathered}$$

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

Using Fubini's Theorem:

$$\begin{gathered} ∫{∫}_R{x}^3{e}^{x^{2}y}dA={∫}_0^4{∫}_{-1}^1{x}^3{e}^{x^{2}y}dydx \\ Evaluationprocedurefortheiteratedintegralweget \\ ={∫}_0^4\left({∫}_{-1}^1{x}^3{e}^{x^{2}y}dy\right)dx \\ ={∫}_0^4{x}^3\left({∫}_{-1}^1{e}^{x^{2}y}dy\right)dx \end{gathered}$$

$$\begin{gathered} Firstwesolve \\ {∫}_{-1}^1{e}^{x^{2}y}dy \\ Put{x}^{2}y=tsowehavedy=\frac{dt}{x^2} \\ Wheny=1thent=x^2 \\ Wheny=-1thent=-x^2 \\ Sointegralbecomes: \\ ={∫}_{-x^2}^{x^2}\frac{e^t}{x^2}dt \\ =\frac{1}{x^2}{∫}_{-x^2}^{x^2}{e}^{t}dt \\ =\frac{1}{x^2}{\left[e^t\right]}_{-x^2}^{x^2} \\ =\frac{e^{x^2}-e^{-x^2}}{x^2} \\ Nowdoubleintegralbecomes: \\ ={∫}_0^4{x}^3\left(\frac{e^{x^2}-e^{-x^2}}{x^2}\right)dx \\ ={∫}_0^{4}x\left(e^{x^2}-e^{-x^2}\right)dx \\ ={∫}_0^{4}x{e}^{x^2}dx-{∫}_0^{4}x{e}^{-x^2}dx \\ =I_1-I_2 \end{gathered}$$

$$\begin{gathered} Solve{I}_{1}weget \\ {∫}_0^{4}x{e}^{x^2}dx \\ Put{x}^2=tsowehavexdx=\frac{dt}{2} \\ whenx=4thent=16 \\ whenx=0thent=0 \\ {I}_{1}becomes{∫}_0^{16}\frac{e^t}{2}dt \\ ={\left[\frac{e^t}{2}\right]}_0^{16} \\ =\frac{e^{16}-1}{2} \\ Solve{I}_{2}weget \\ Put-x^2=tsowehavexdx=-\frac{dt}{2} \\ whenx=4thent=-16 \\ whenx=1thent=0 \\ {I}_{1}becomes{∫}_0^{-16}-\frac{e^t}{2}dt \\ ={\left[-\frac{e^t}{2}\right]}_0^{-16} \\ =-\frac{e^{-16}-1}{2} \\ =\frac{1-e^{-16}}{2} \\ Sowehave \\ {I}_1-I_2=\frac{e^{16}-1}{2}-\frac{1-e^{-16}}{2} \\ =\frac{e^{16}-1}{2}-\frac{e^{16}-1}{2e^{16}} \\ =\frac{e^{32}-2e^{16}+1}{2e^{16}} \end{gathered}$$