Question

The area of the region bounded by $y-x=2$ and $x^2=y$ is equal to

• A. $\frac{2}{3}$
• B. $\frac{4}{3}$
• C. $\frac{16}{3}$
• D. $\frac{9}{2}$

Answer 1

The correct option is D $\frac{9}{2}$


On solving equations $x^2=y$
and $y–x=2$ we get
$\Rightarrow x^2=2+x$
$\Rightarrow x^2-x-2=0$
$\Rightarrow (x-2)(x+1)=0$
$\Rightarrow x=-1,2$
$\therefore$ Required area $=\underset{-1}{\overset{2}{\int }}(x+2-x^2)dx$
$={[\frac{x^2}{2}+2x-\frac{x^3}{3}]}_{-1}^2$
$=(2+4-\frac{8}{3}-(\frac{1}{2}-2+\frac{1}{3}))$
$=\frac{9}{2}$ sq. units