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

Given that

${y^2} = 9x$ and $y = 3x$
Solving ${y^2} = 3(3x) = 3y$
$\Rightarrow y = 0$ or 3

When $y = 0,x = 0$ and when $y = 3,x = 1$

So points of intersection are (0,0) and (1,3) Graphs of parabola$\;{y^2} = 9x$ and line $y = 3x$are as shown in the adjacent figure.

From the figure, Area of shaded region
$A = \int_0^1 {(\sqrt {9x} - 3x)} dx$

$= 3\int_0^1 {{x^{1/2}}} dx - 3\int_0^1 x dx$
$= 3\left[ {\frac{{{x^{3/2}}}}{{3/2}}} \right]_0^1 - 3\left[ {\frac{{{x^2}}}{2}} \right]_0^1$

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

$= 2 - \frac{1}{2} = - \frac{1}{2}$ (anits.