Q. 46 - Calculus

Question

Evaluate each of the double integrals in Exercises 37–54 as iterated integrals. 

$$\begin{gathered} \int {\int }_R{x}^3{e}^{x^{2}y}dA, \\ where R=\left\{\right(x,y\left)\right| 0\le x\le 4 and -1\le y\le 1\} \end{gathered}$$

Step-by-Step Solution

Answer

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

Step 1. Given Information:

Given double integrals : 

$$\begin{gathered} \int {\int }_R{x}^3{e}^{x^{2}y}dA, \\ where R=\left\{\right(x,y\left)\right| 0\le x\le 4 and -1\le y\le 1\} \end{gathered}$$

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

Step 2. Solution:

Using Fubini's Theorem:

$$\begin{gathered} \int {\int }_R{x}^3{e}^{x^{2}y}dA={\int }_0^4{\int }_{-1}^1{x}^3{e}^{x^{2}y} dy dx \\ Evaluation procedure for the iterated integral we get \\ ={\int }_0^4\left({\int }_{-1}^1{x}^3{e}^{x^{2}y} dy\right) dx \\ ={\int }_0^4{x}^3\left({\int }_{-1}^1{e}^{x^{2}y} dy\right) dx \end{gathered}$$

$$\begin{gathered} First we solve \\ {\int }_{-1}^1{e}^{x^{2}y} dy \\ Put x^{2}y=t so we have dy=\frac{dt}{x^2} \\ When y=1 then t=x^2 \\ When y=-1 then t=-x^2 \\ So integral becomes: \\ ={\int }_{-x^2}^{x^2}\frac{e^t}{x^2} dt \\ =\frac{1}{x^2}{\int }_{-x^2}^{x^2}{e}^t dt \\ =\frac{1}{x^2}{\left[e^t\right]}_{-x^2}^{x^2} \\ =\frac{e^{x^2}-e^{-x^2}}{x^2} \\ Now double integral becomes: \\ ={\int }_0^4{x}^3\left(\frac{e^{x^2}-e^{-x^2}}{x^2}\right)dx \\ ={\int }_0^{4}x\left(e^{x^2}-e^{-x^2}\right)dx \\ ={\int }_0^{4}x{e}^{x^2} dx-{\int }_0^{4}x{e}^{-x^2}dx \\ =I_1-I_2 \end{gathered}$$

$$\begin{gathered} Solve I_{1}we get \\ {\int }_0^{4}x{e}^{x^2} dx \\ Put x^2=t so we have x dx=\frac{dt}{2} \\ when x=4 then t=16 \\ when x=0 then t=0 \\ {I}_1 becomes {\int }_0^{16}\frac{e^t}{2} dt \\ ={\left[\frac{e^t}{2}\right]}_0^{16} \\ =\frac{e^{16}-1}{2} \\ Solve I_{2 }we get \\ Put -x^2=t so we have x dx=-\frac{dt}{2} \\ when x=4 then t=-16 \\ when x=1 then t=0 \\ {I}_1 becomes {\int }_0^{-16}-\frac{e^t}{2} dt \\ ={\left[-\frac{e^t}{2}\right]}_0^{-16} \\ =-\frac{e^{-16}-1}{2} \\ =\frac{1-e^{-16}}{2} \\ So we have \\ {I}_1-I_2=\frac{e^{16}-1}{2}-\frac{1-e^{-16}}{2} \\ =\frac{e^{16}-1}{2}-\frac{e^{16}-1}{2e^{16}} \\ =\frac{e^{32}-2e^{16}+1}{2e^{16}} \end{gathered}$$