Question

The value of integral over the region in positive quadrant for which $x+y\le 1$ is ______

• A. 0.16

Answer 1

The correct option is A 0.16
The region of integration R is the area bounded by the straight lines $x+y=1$, and the co-ordinate axes i.e it is bounded by the lines.
$x=0,x=1-y,y=0,andy=1$

Integrating first over the horizontal strip $PQ$ i.e w.r.t from the point $P(x=0)$ to $Q(x=1-y)$ and then w.r.t y from $y=0$ to $y=1$ we get

$\int {\int }_R(x^2+y^2)dxdy={\int }_0^1{\int }_0^{1-y}(x^2+y^2)dxdy$

$={\int }_0^1{[\frac{x^3}{3}+x{y}^2]}_0^{1-y}dy={\int }_0^1[\frac{(1-y{)}^3}{3}+(1-y\left)y^2\right]dy$

$={[\frac{-(1-y{)}^4}{12}-\frac{y^3}{3}-\frac{y^4}{4}]}_0^1=(\frac{1}{3}-\frac{1}{4})-(-\frac{1}{12})=\frac{1}{6}=0.1667$