$$ \int _ { 0 } ^ { 2 } \int _ { \frac { 1 } { y } } ^ { 1 } x ^ { 2 } e ^ { x } d x d y $$

Question

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We start with the iterated integral

$$
\int_{0}^{2} \int_{\frac{1}{y}}^{1} x^{2} e^{x}\,dx\,dy.
$$

Note that for $$y \in (0,1)$$ the lower limit $$\frac{1}{y}$$ is greater than 1 so the region where $$x$$ varies is empty. In effect, the integration over $$y$$ from 0 to 1 gives zero and the region of integration is actually

$$
D=\{(x,y): 1\le y\le 2,\; \frac{1}{y}\le x\le 1\}.
$$

A good strategy is to change the order of integration. In the region $$D$$, the inequality

$$
\frac{1}{y}\le x\le 1
$$

can be rewritten as

$$
y\le \frac{1}{x}\quad \text{with}\quad  y\ge 1.
$$

Also, when $$y=2$$ we have $$x\ge \frac{1}{2}$$. Hence, upon switching the order, the variable $$x$$ runs from $$\frac{1}{2}$$ to $$1$$. For a fixed $$x$$, the variable $$y$$ runs from $$y=1$$ up to $$y=\frac{1}{x}$$ (since $$\frac{1}{x} \le 2$$ for $$x\in \left[\frac{1}{2},1\right]$$). Thus the integral becomes

$$
\int_{x=\frac{1}{2}}^{1} \int_{y=1}^{\frac{1}{x}} x^{2} e^{x}\,dy\,dx.
$$

Since the integrand $$x^2e^x$$ is independent of $$y$$, we first integrate with respect to $$y$$:

$$
\int_{y=1}^{\frac{1}{x}} dy = \frac{1}{x} - 1.
$$

Thus, the integral reduces to

$$
\int_{\frac{1}{2}}^{1} x^{2} e^{x}\Bigl(\frac{1}{x}-1\Bigr)dx
=\int_{\frac{1}{2}}^{1} x^{2}e^{x}\frac{1-x}{x}dx
=\int_{\frac{1}{2}}^{1} x e^{x}(1-x)dx.
$$

We now need to compute

$$
I=\int_{\frac{1}{2}}^{1} x e^{x}(1-x)dx.
$$

Expanding the integrand gives

$$
I=\int_{\frac{1}{2}}^{1} \Bigl(xe^{x} - x^{2}e^{x}\Bigr)dx
= \int_{\frac{1}{2}}^{1} xe^{x}\,dx - \int_{\frac{1}{2}}^{1} x^{2}e^{x}\,dx.
$$

Let us denote

$$
I_{1}=\int xe^{x}\,dx,\qquad I_{2}=\int x^{2}e^{x}\,dx.
$$

### Evaluating $$I_{1}$$

Use integration by parts: let
$$
u=x,\quad dv=e^x\,dx,\quad du=dx,\quad v=e^x.
$$
Then

$$
I_{1} =xe^{x}-\int e^{x}dx = xe^{x}-e^{x}= e^{x}(x-1).
$$

Thus,

$$
\int_{\frac{1}{2}}^{1} xe^{x}\,dx = \Bigl[e^{x}(x-1)\Bigr]_{\frac{1}{2}}^{1}.
$$

Evaluating at the bounds:
- At $$x=1$$: $$e^{1}(1-1)=0$$.
- At $$x=\frac{1}{2}$$: $$e^{\frac{1}{2}}\Bigl(\frac{1}{2}-1\Bigr)= -\frac{1}{2}e^{\frac{1}{2}}$$.

Thus,

$$
\int_{\frac{1}{2}}^{1} xe^{x}\,dx = 0 - \Bigl(-\frac{1}{2}e^{\frac{1}{2}}\Bigr)
=\frac{1}{2}e^{\frac{1}{2}}.
$$

### Evaluating $$I_{2}$$

Again, apply integration by parts: let
$$
u=x^{2},\quad dv=e^x\,dx,\quad du=2x\,dx,\quad v=e^x.
$$
Then

$$
I_{2}=x^{2}e^{x} - \int 2xe^{x}dx.
$$

We have already computed $$\int xe^{x}\,dx = e^{x}(x-1)$$. Therefore,

$$
I_{2}= x^{2}e^{x} - 2e^{x}(x-1)
= e^{x}\bigl(x^{2}-2x+2\bigr).
$$

Thus,

$$
\int_{\frac{1}{2}}^{1} x^{2}e^{x}dx = \Bigl[e^{x}\bigl(x^{2}-2x+2\bigr)\Bigr]_{\frac{1}{2}}^{1}.
$$

Evaluating at the bounds:
- At $$x=1$$: 
  $$
  e^{1}(1-2+2)=e(1)= e.
  $$
- At $$x=\frac{1}{2}$$:
  $$
  e^{\frac{1}{2}}\Biggl[\left(\frac{1}{2}\right)^2 -2\left(\frac{1}{2}\right)+2\Biggr]
  = e^{\frac{1}{2}}\Bigl(\frac{1}{4}-1+2\Bigr)
  = e^{\frac{1}{2}}\left(\frac{1}{4}+1\right)
  =\frac{5}{4}e^{\frac{1}{2}}.
  $$

Thus,

$$
\int_{\frac{1}{2}}^{1} x^{2}e^{x}dx = e - \frac{5}{4}e^{\frac{1}{2}}.
$$

### Combining the results

Recall

$$
I = \int_{\frac{1}{2}}^{1} xe^{x}\,dx - \int_{\frac{1}{2}}^{1} x^{2}e^{x}\,dx.
$$

Substitute our results:

$$
I = \frac{1}{2}e^{\frac{1}{2}} - \Bigl(e - \frac{5}{4}e^{\frac{1}{2}}\Bigr)
= \frac{1}{2}e^{\frac{1}{2}} - e + \frac{5}{4}e^{\frac{1}{2}}
= \left(\frac{1}{2}+\frac{5}{4}\right)e^{\frac{1}{2}}-e.
$$

Compute the coefficient:

$$
\frac{1}{2}+\frac{5}{4}=\frac{2}{4}+\frac{5}{4}=\frac{7}{4}.
$$

Thus, the final answer is

$$
\frac{7}{4}e^{\frac{1}{2}}-e.
$$
The value of the integral is $$ \frac{7}{4}e^{\frac{1}{2}} - e $$.

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