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

Marks : 6

Year : 2013

1) The limit of x is 0 to 1

i.e. $x=0$ and $x=1$

2.1) The limit of y are $y=x$ let $x=0$ then $y=0$

When $x=1$ then $y=1$

$\therefore y=x$ is a line passing from $(0,0)$ and $(1,1)$

2.2) Upper limit of y is $y=2-x$

i.e. when $x=0 y=2$

When $x=2 y=0$

i.e. line passing from $(0,2)$ and $(2,0).$

Now changing the order we take horizontal strip now the outer limit will be y and will be from 0 to 2.

But the inner upper limit is not constant and changes at $y=1$

$\therefore Region 1$

Limit of $x: x=0 x=y$

Limits of $y: y=0 $ to $ y=1$

Region 2:

Limits of $x: x=0 $ to $ x=2-y$

Limits of $y: y=1 $ to $ y=2$

$$\therefore I=\int\limits_0^1\int_0^y\dfrac xy dxdy+\int_1^2\int_0^{2-y}\dfrac xy dxdy$$

Integrating w.r.t.x.

$=\int\limits_0^1\dfrac 1y\Bigg[\dfrac {x^2}2\Bigg]^y_0dy+\int_1^2\dfrac 1y\Bigg[\dfrac {x^2}2\Bigg]_0^{2-y}dy\\ =\int\limits_0^1\dfrac 1y.\dfrac {y^2}2dy +\int\limits_1^2\dfrac 1y \dfrac {(2-y)^2}2.dy\\ =\int\limits_0^1\dfrac y2 dy+ \dfrac 12 \int\limits_1^2\Bigg(\dfrac 4y-4+y\Bigg)dy \\ I=\Bigg[\dfrac {y^2}4\Bigg]_0^1 +\dfrac 12\Bigg[4\log y-4y+\dfrac {y^2}2\Bigg]_1^2 \\ =\dfrac 14+\dfrac 12\Bigg[ 4\log 2-4(2)+\dfrac 42- 4\log 1+4(1)-\dfrac 12\Bigg]\\ =\dfrac 14+\dfrac 12\Bigg[4\log 2-8+2+4-\dfrac 12\Bigg]\\ =\dfrac 14+\dfrac 12\Bigg[4\log 2-\dfrac 52\Bigg]\\ =\dfrac 14+2\log 2-\dfrac 54\\ =2\log 2-1\\ =\log 4-1 $