Application of Integrals — Class 12 Maths Solution

We have, ${y^2} = 9x$ and $y = x$
Solving ${y^2} = 9y$
$\Rightarrow y = 0$ or 9
When $y = 0,x = 0$ and when $y = 9,x = 9$

So points of intersection are (0,0) and (9,9)
Graph of parabola ${y^2} = 9x$ and $y = x$ are as shown in the following figure.

From the figure, area of shaded region
$A = \int_0^9 {(\sqrt {9x} - x)} dx$
$= 3\int_0^9 {{x^{1/2}}} dx - \int_0^9 x dx$

$= 3\left[ {\frac{{{x^{3/2}}}}{{3/2}}} \right]_0^9 - \left[ {\frac{{{x^2}}}{2}} \right]_0^9$

$= 3\left( {\frac{2}{3}27 - 0} \right) - \left( {\frac{{81}}{2} - 0} \right) = 54 - \frac{{81}}{2} = \frac{{27}}{2}8q$ units