Mumbai University > First Year Engineering > sem 2 > Applied Maths 2

Marks : 6

Year : 2014

The limit of y is

$y=0$ straight line

$y=x$ straight line

The limit of x is $x=0 \space to\space x= \infty$

On changing order

1) Outer limit y

$y=0 \space to\space y= \infty$

2) Inner limit y

$x=0 \space to\space x= \infty$

$ \therefore I=\int\limits_{y=0}^{y=\infty}\int\limits_{x=y}^{x=\infty}xe^{\frac {-x^2}y}dxdy \\ Put \space \dfrac {x^2}y=U \space Diff \space w.r.t.x \\ \therefore \dfrac {2x}ydx=dU \\ when \space x=y,\therefore U=y \\ when\space x=\infty, U=\infty \\ \therefore I=\int\limits_{0}^{\infty} \int\limits_{y}^{\infty} e^{-U}.\dfrac {ydU}2dy \\ =\int\limits_{0}^{\infty}\dfrac y2\Big[\dfrac {e^{-U}}{-1}\Big]^{\infty}_ydy \\ =\int\limits_0^{\infty}\dfrac y2(-1[0-e^{-y}])dy \\ =\dfrac 12\int\limits_0^{\infty}ye^{-y}dy \\ \text{Integrating by parts,} \\ =\dfrac 12\Bigg[y\dfrac {e^{-y}}{-1}-\int 1\dfrac {e^{-y}}{-1}dy\Bigg]_0^{\infty} \\ =\dfrac 12[-ye^{-y}-e^{-y}]_0^{\infty} \\ =\dfrac 12[-0-0]-[-0-1] \\ =\dfrac 12$