Question

If the area of the closed figure bounded by the curves $y=\sqrt{x},y=\sqrt{4-3x}$ and $y=0$ is $\frac{\alpha }{9},(\alpha \in N)$ then $\alpha$ is

Answer 1

Required area
$=\underset{0}{\overset{1}{\int }}\sqrt{x}dx+\underset{1}{\overset{4/3}{\int }}\sqrt{4-3x}dx=\frac{8}{9}\text{sq. units}$
$\Rightarrow \alpha =8$