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

Marks : 6

Year : DEC 2013

Integrating w.r.t Z

$\therefore I= \int\limits_0^1 \int\limits^1_{y^2} x[z]_0^{1-x} dy \space dx \\ = \int\limits_0^1\int\limits^1_{y^2} x [1-x] dx\space dy$

Integrating w.r.t x

$I=\int\limits_0^1 \int\limits^1_{y^2}x-x^2 dx\space dy\\ = \int\limits_0^1\Bigg[\dfrac {x^2}2-\dfrac {x^3}3\Bigg]^1_{y^2}dy\\ =\int\limits^1_0\Bigg(\dfrac 12-\dfrac 13\Bigg)-\Bigg(\dfrac {y^4}2-\dfrac {y^6}3\Bigg)dy\\ =\int\limits^1_0\dfrac 16-\dfrac {y^4}2+\dfrac {y^6}3dy $

Integrating w.r.t y

$I=\Bigg[\dfrac 16y-\dfrac 12\dfrac {y^5}5+\dfrac 13\dfrac {y^7}7\Bigg]_0^1\\ =\Bigg[\dfrac 16-\dfrac 1{10}+\dfrac 1{21}\Bigg]-[0]\\ =\dfrac 4{35}$