Question

The area of the closed figure bounded by the curves $y=\sqrt{x},y=\sqrt{4-3x}$ and $y=0$ is:

• A. $\frac{4}{9}\text{sq. units}$
• B. $\frac{8}{9}\text{sq. units}$
• C. $\frac{16}{9}\text{sq. units}$
• D. $\frac{5}{9}\text{sq. units}$

Answer 1

The correct option is B

$\frac{8}{9}\text{sq. units}$

Solving both the curve, we get
$x=4-3x$
(point of intersection) $\Rightarrow x=1$

So, $A={\int }_0^1(\frac{4-y^2}{3}-y^2)dy=\frac{8}{9}$

Alternative:
${\int }_0^1\sqrt{x}dx+{\int }_1^{4/3}\sqrt{4-3x}dx=\frac{8}{9}\text{sq. units}$