I am supposed to evaluate a double integral using change of variables. The integral is: $ \int_{0}^{1}\int_{0}^{x}\frac{(x+y)e^{(x+y)}}{{x^{2}}}dydx $
I am given these equations to switch between variables:
$ u=\frac{y}{x}$ and $v=x+y$
Calculations:
The first thing to calculate is the Jacobian:
$dxdy$ = $\begin{bmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y}\\ \frac{\partial v}{\partial x}& \frac{\partial v}{\partial y} \end{bmatrix}$
$\frac{\partial v}{\partial x}=\frac{\partial v}{\partial y}=1$
$\frac{\partial u}{\partial x}=\frac{-y}{x^{2}}$ and $\frac{\partial u}{\partial y}=\frac{1}{x}$
Therefore the Jacobian = $dxdy$ = $\frac{-y}{x^{2}}-\frac{1}{x}$
= $\frac{-y-x}{x^{2}}$
and the integral becomes $ \int_{0}^{1}\int_{0}^{x}\frac{(x+y)e^{(x+y)}}{{x^{2}}}(\frac{-y-x}{x^{2}}) dudv $
Since $v=x+y$,
the integral becomes $ \int_{0}^{1}\int_{0}^{x}\frac{(v)e^{(v)}(-v)}{{x^{2}}}dudv $
$ \int_{0}^{1}\int_{0}^{x}\frac{(-v^{2})e^{(v)}}{{x^{2}}}dudv $
At this point I am confused how to substitute for the x in the integral bound and the denominator as well as changing the boundaries in the integrals. I know I want the entire integral in terms of u and v. I was thinking about using $x=v-y$ but that leaves it in terms I don't want.
The answer is $e^{2}-e-1$
Thanks for the help.