Discovering Statistics

A den for Learning


JAM 2022 [ 51-60 ]

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=-p

Determining 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

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