Question

Find the area of the closed figure bounded by the following curves
$y=$ $\sqrt{x},y=\sqrt{4-3x}$, y = 0.

• A. $\frac{8}{9}$
• B. $\frac{7}{9}$
• C. $\frac{5}{9}$
• D. $\frac{1}{3}$

Answer 1

The correct option is A $\frac{8}{9}$

Given curve, $y=\sqrt{x},y=\sqrt{4-3x},y=0$

$y>0,x\ge 0,y-3x\ge 0$ (basic square root conditions)

$\Rightarrow x\le 4/3$ __(1)

Now, $y=\sqrt{4-3x}\Rightarrow y^2=4-3x$

$x=\frac{4-y^2}{3}=\frac{(2-y)(2+y)}{3}$

It represents a parabola cutting y-axis at $(0,2),(0,-2)$ & x-axis at $(4/3,0)$

Ref. image

Point of intersection

$\sqrt{4-3x}=\sqrt{x}$

$\Rightarrow 4-3x=x$

$x=1\Rightarrow y=\sqrt{1}=1$

$(1,1)$ is req. point

Area bounded by curves (shaded)

$={\int }_0^1\sqrt{x}dx+{\int }_1^{4/3}\sqrt{4-3x}dx$

$=\frac{2}{3}{\left[x^{3/2}\right]}_0^1+{[\frac{2}{3}\times {\frac{(4-3x)}{(-3)}}^{3/2}]}_1^{4/3}$

$=\frac{2}{3}(1-0)+\left(\frac{-2}{9}\right)(0-1)$

$=\frac{2}{3}+\frac{2}{9}=\frac{8}{9}$

$\therefore$ option A is correct.