The value of
\int_{0}^2\int_{0}^{2-x} (x+y)^2e^{\frac{2y}{x+y}}dydx equals _______________ (round off to 2 decimal places)
\textit{Transform, } p= x+y \\ z= \frac{y}{x+y}= \frac{y}{p}The inverse transform of which would be,
y=zp \\
x= p-zp \\
\textit{So,}
\frac{d}{dp}y =z , \frac{d}{dp}x = 1-z \\
\frac{d}{dz}y =p , \frac{d}{dp}z = -p \\
J= \left| \begin{matrix} z & 1-z \\
p & -p
\end{matrix} \right| = -zp-p+zp=-pDetermining the conditional ranges,
0< y< 2-x \Rightarrow 0< zp< 2-p+zp \Rightarrow z,p>0 , p<2 \\ 0 < x < 2 \Rightarrow 0< p(1-z) < 2 \Rightarrow 0 < z < 1
Thus the integration may be written as:
\int_{0}^1\int_{0}^{2} p^3e^{2z} dpdz = \left[ \frac{p^4}{4} \right]_{0}^{2} * \left[ \frac{e^{2z}}{2} \right]_{0}^{1} = 2(e^{2} -1)=12.778