The area (in sq. units) of the region bounded by $y-x=2$ and $x^2=y$ is equal to
- $\frac{2}{3}$
- $\frac{4}{3}$
- $\frac{9}{2}$
- $\frac{16}{3}$
Solution
$\begin{array}{ll} & y-x=2 \text { and } x^2=y \\ \therefore \quad & x^2-x-2=0 \\ \therefore \quad & (x-2)(x+1)=0 \\ \therefore \quad & x=2 \text { or } x=-1 \\ \therefore \quad & y=4, y=1 \end{array}$ $\therefore \quad$ Points of intersection of the two curves are - $(2,4)$ and $(-1,1)$ $\begin{aligned} \text { Required area } & =\int_{-1}^2(2+x)-x^2 d x \\ & =2[x]_{-1}^2+\frac{1}{2}\left[x^2\right]_{-1}^2-\frac{1}{3}\left[x^3\right]_{-1}^2 \\ & =6+\frac{3}{2}-3=\frac{9}{2} \text { sq. units } \end{aligned}$
Asked in: MHT CET 2024 (11 May Shift 2)