Question

Find the area of the region bounded by curves $y^2=9x,y=3x$

Answer 1

We have $y^2=9x\cdots \left(1\right)$
& $y=3x\cdots \left(2\right)$

Substitute the value of $y$ from equation (2) to equation (1)
$\Rightarrow 9x^2=9x$
$\Rightarrow x(x-1)=0$
$\Rightarrow x=0,1$

When $x=0,y=0$ & when $x=1,y=3$
[using (2)]

Hence$(0,0)\text{and}(1,3)$ are the points of intersection.

So, the area bounded by curves is shaded in the diagram below:


Area $={\int }_{x_1}^{x_2}(y_2-y_1)dx$

Area $={\int }_0^1(\sqrt{9x}-3x)dx$
$[\because x\text{varies from}0to1]$

$={\int }_0^1(\sqrt{9x}dx-{\int }_0^{1}3xdx$

$=3{\left[\frac{2}{3}{x}^{\frac{3}{2}}\right]}_0^1-3{\left[\frac{x^2}{2}\right]}_0^1$ $[\because {\int }_a^b{x}^{n}dx={\left[\frac{x^{n+1}}{n+1}\right]}_a^b]$

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

$2-\frac{3}{2}-\frac{1}{2}sq.units$

Hence the required area is $\frac{1}{2}sq.units$.