Question

In Exercises 59–62, evaluate the double integral over the specified region.

$\int {\int }_{\Omega }x{e}^{x^3}dA$

Short answer

Value of the integral $\int {\int }_{\Omega }x{e}^{x^3}dA$over the rectangular region is,

${\int }_0^2{\int }_y^{2}x{e}^{x^3}dxdy=\frac{e^8-1}{3}$.

Step-by-step solution

Step 1. Given information.

We have given integral,

$\int {\int }_{\Omega }x{e}^{x^3}dA$

over the triangular region with vertices (0, 0), (2, 0) and (2, 2).

Step 2. Explanation.

Drawing the rectangular region,

From the above region,

$dA=dxdy$

and $y\le x\le 2,0\le y\le 2$

Hence,

${\int }_0^2{\int }_y^{2}x{e}^{x}dxdy$

Changing the order of integration,

${\int }_0^2{\int }_y^{2}x{e}^{x^3}dxdy\hspace{0.17em}\hspace{0.17em}={\int }_0^2{\int }_0^{x}x{e}^{x^3}dydx$

Evaluating the integral,

${\int }_0^2{\int }_0^{x}x{e}^{x^3}dydx={\int }_0^2{x}^2{e}^{x^3}dx$

Substituting $$\begin{gathered} x^3=u \\ \Rightarrow du=3x^{2}dx \\ \Rightarrow x^{2}dx=\frac{1}{3}du \end{gathered}$$

${\int }_0^2{x}^2{e}^{x^3}dx={\int }_0^8\frac{e^u}{3}du$

=$\frac{1}{3}{\left[e^u\right]}_0^8$

=$\frac{e^8-1}{3}$

Step 3. Conclusion.

Hence, value of the integral $\int {\int }_{\Omega }x{e}^{x^3}dA$over the rectangular region is,

${\int }_0^2{\int }_y^{2}x{e}^{x^3}dxdy=\frac{e^8-1}{3}$.