The area (in sq unit) of the region bounded by the curves $2 x=y^2-1$ and $x=0$ is
- $\frac{1}{3}$
- $\frac{2}{3}$
- $1$
- $2$
Solution

$\therefore$ Required area $=\int_{-1}^1 x d y$ $\begin{aligned} & =2 \int_0^1 \frac{y^2-1}{2} d y \\ & =\left|\left[\frac{y^3}{3}-y\right]_0^1\right| \\ & =\left|\left[\frac{1}{3}-1\right]\right|=\frac{2}{3} \text { sq unit }\end{aligned}$
Asked in: AP EAMCET 2008