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 \( x+y=3 \)

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:** \[ \boxed{\frac{1}{4}\left(e^{6}-1\right)-\frac{3}{2}} \]