The area (in square units) of the region bounded by $x^2=8 y, x=4$ and $X$-axis, is
- $\frac{2}{3}$
- $\frac{4}{3}$
- $\frac{8}{3}$
- $\frac{10}{3}$
Solution

On solving Eqs. (i) and (ii), we get $(4,2)$ The required area $=\int_0^4 y d x=\int_0^4 \frac{x^2}{8} d x=\left[\frac{x^3}{24}\right]_0^4=\frac{64}{24}=\frac{8}{3}$ The required area $=\frac{8}{3}$ sq units
Asked in: AP EAMCET 2001