5. Use an appropriate change of variables to evaluate the integral $$ \iint_{R}(x+y) \mathrm{e}^{x^{2}-y^{2}} d A $$ where $$ R $$ is the rectangle enclosed by the lines $$ x-y=0, x-y=2, x+y=0 $$, and $$ x+y=3 $$

Question

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We begin by considering the integral

$$
I=\iint_R (x+y) \, e^{x^2-y^2}\,dA,
$$

with the region $$ R $$ bounded by the lines

$$
x-y=0,\quad x-y=2,\quad x+y=0,\quad x+y=3.
$$

Because the boundaries are given in terms of $$ x-y $$ and $$ x+y $$, an appropriate change of variables is

$$
u=x+y,\quad v=x-y.
$$

This transformation will also simplify the integrand.

---

**Step 1. Express $$ x $$ and $$ y $$ in terms of $$ u $$ and $$ v $$**

Solve for $$ x $$ and $$ y $$:

$$
x=\frac{u+v}{2},\quad y=\frac{u-v}{2}.
$$

---

**Step 2. Compute the Jacobian**

The Jacobian $$ J $$ is determined by

$$
J=\begin{vmatrix}
\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v}\[1mm]
\frac{\partial y}{\partial u} & \frac{\partial y}{\partial v}
\end{vmatrix}.
$$

We have

$$
\frac{\partial x}{\partial u}=\frac{1}{2},\quad \frac{\partial x}{\partial v}=\frac{1}{2},\quad \frac{\partial y}{\partial u}=\frac{1}{2},\quad \frac{\partial y}{\partial v}=-\frac{1}{2}.
$$

Thus,

$$
J=\begin{vmatrix}
\frac{1}{2} & \frac{1}{2}\[1mm]
\frac{1}{2} & -\frac{1}{2}
\end{vmatrix}=\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)-\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)
=-\frac{1}{4}-\frac{1}{4}=-\frac{1}{2}.
$$

Taking the absolute value, we have

$$
|J|=\frac{1}{2}\quad \Longrightarrow \quad dA = \frac{1}{2}\,du\,dv.
$$

---

**Step 3. Rewrite the integrand**

The integrand consists of two parts:

1. The factor $$ x+y $$ becomes
   $$
   x+y=u.
   $$

2. The exponent $$ x^2-y^2 $$. We calculate:
   $$
   x^2-y^2=\left(\frac{u+v}{2}\right)^2-\left(\frac{u-v}{2}\right)^2.
   $$
   Expanding both squares:
   $$
   \left(\frac{u+v}{2}\right)^2=\frac{u^2+2uv+v^2}{4},\quad \left(\frac{u-v}{2}\right)^2=\frac{u^2-2uv+v^2}{4}.
   $$
   Taking their difference:
   $$
   x^2-y^2=\frac{(u^2+2uv+v^2)-(u^2-2uv+v^2)}{4} =\frac{4uv}{4}=uv.
   $$
   Thus,
   $$
   e^{x^2-y^2}=e^{uv}.
   $$

So the integrand in the new variables becomes

$$
(x+y)e^{x^2-y^2}=u\,e^{uv}.
$$

---

**Step 4. Transform the region $$ R $$**

The region $$ R $$ is given by:
- $$ x-y=0 $$ becomes $$ v=0 $$.
- $$ x-y=2 $$ becomes $$ v=2 $$.
- $$ x+y=0 $$ becomes $$ u=0 $$.
- $$ x+y=3 $$ becomes $$ u=3 $$.

Thus, in the $$(u,v)$$ plane, the region is the rectangle:

$$
0\leq u\leq 3,\quad 0\leq v\leq 2.
$$

---

**Step 5. Write the new integral**

Substituting everything, we have

$$
I=\iint_R (x+y)e^{x^2-y^2}\,dA
=\int_{v=0}^{2}\int_{u=0}^{3} u\, e^{uv} \left(\frac{1}{2}\,du\,dv\right)
=\frac{1}{2}\int_{v=0}^{2}\int_{u=0}^{3} u\,e^{uv}\,du\,dv.
$$

---

**Step 6. Change the order of integration**

It is easier to integrate with respect to $$ v $$ first. Rewrite the integral by switching the order:

$$
I=\frac{1}{2}\int_{u=0}^{3}\int_{v=0}^{2} u\,e^{uv}\,dv\,du.
$$

For a fixed $$ u $$, the inner integral with respect to $$ v $$ is:

$$
\int_{v=0}^{2} e^{uv}\,dv.
$$

Since $$ u $$ is treated as constant, we have:

$$
\int_{v=0}^{2} e^{uv}\,dv = \frac{1}{u}\left(e^{u\cdot 2}-1\right)
=\frac{1}{u}\left(e^{2u}-1\right).
$$

Thus, the integral becomes

$$
I=\frac{1}{2}\int_{u=0}^{3} u\left[\frac{1}{u}\left(e^{2u}-1\right)\right] du
=\frac{1}{2}\int_{u=0}^{3}\left(e^{2u}-1\right)\,du.
$$

---

**Step 7. Evaluate the integral with respect to $$ u $$**

Break the integral into two parts:

$$
I=\frac{1}{2}\left(\int_{u=0}^{3}e^{2u}\,du - \int_{u=0}^{3}1\,du\right).
$$

1. For the first integral:
   $$
   \int e^{2u}\,du = \frac{1}{2}e^{2u},
   $$
   so
   $$
   \int_{0}^{3} e^{2u}\,du=\frac{1}{2}\left(e^{6}-1\right).
   $$

2. For the second integral:
   $$
   \int_{0}^{3}1\,du=3.
   $$

Thus,

$$
I=\frac{1}{2}\left[\frac{1}{2}\left(e^{6}-1\right)-3\right]
=\frac{1}{4}\left(e^{6}-1\right)-\frac{3}{2}.
$$

---

**Final Answer:**

$$
\boxed{\frac{1}{4}\left(e^{6}-1\right)-\frac{3}{2}}
$$
The integral evaluates to $$ \frac{1}{4}(e^{6} - 1) - \frac{3}{2} $$.

Use an appropriate change of variables to evaluate the integral

where is the rectangle enclosed by the lines , and

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Extra Insights

Next, we compute the Jacobian of the transformation. The inverse transformations are:

Calculating the Jacobian gives us:
[
\frac{\partial(x,y)}{\partial(u,v)} =

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