Question

$\mathbf{\iint }{\mathit{x}}^{\mathbf{2}}{\mathit{e}}^{{\mathbf{x}}^{\mathbf{2}}\mathbf{y}}\mathit{d}\mathit{x}\mathit{d}\mathit{y}$over the area bounded by$\mathit{y}\mathbf{=}{\mathit{x}}^{\mathbf{-}\mathbf{x}}\mathbf{,}\mathit{y}\mathbf{=}{\mathit{x}}^{\mathbf{-}\mathbf{2}}$and$\mathit{x}\mathbf{=}\mathit{I}\mathit{n}\mathbf{4}$

Short answer

The required solution is $4-eIn4$.

Step-by-step solution

Definition of double integral

The double integral of $\mathit{f}\left(x,y\right)$over the area Ain the$\left(x,y\right)$plane as the limit of this sum, and we write it as${\mathbf{\iint }}_{\mathbf{A}}\mathit{f}\left(x,y\right)\mathit{d}\mathit{x}\mathit{d}\mathit{y}$.

Drawing the area bounded by the curve

The area bounded by the graph $y=\frac{1}{x}\text{'}y=\frac{1}{x_2}$and $x=In4$.

Intersecting points identification

From the figure the curves$\frac{1}{x}$ and$\frac{1}{x^2}$ intersect at$\left(x,y\right)=\left(1,1\right)$.

From the figure the curves and intersect at

Thus the integral is:

$$\begin{gathered} {\iint }_A{x}^2{e}^{y{x}^2}dxdy={\int }_{x=1}^{In4}\left[{\int }_{y=1/x^2}^{1/x}{e}^{y{x}^2}d\left(y{x}^2\right)\right]dx \\ ={\int }_1^{In4}\left(e^x-e\right)dx \\ ={\overline{)\left(e^x-xe\right)}}_1^{In4} \\ =4-eIn4 \end{gathered}$$

Therefore, the value is$4-eIn4$.