Find the area of the region included between $y^2 = 9x$ and $y = x$.

$\frac{27}{2}$

The correct answer is Option (2) → $\frac{27}{2}$

We have, $y^2 = 9x$ and $y = x$

On solving both equations, we get

$x^2 = 9x$

$\Rightarrow x^2 - 9x = 0$

$\Rightarrow x(x - 9) = 0$

$\Rightarrow x = 0, 9$

$∴$ Area of shaded region $=$ Area under the curve, $(y^2 = 9x)$ $-$ area under the line, $(y = x)$

$= \int_{0}^{9} \sqrt{9x} \, dx - \int_{0}^{9} x \, dx$

$= \int_{0}^{9} 3x^{1/2} \, dx - \int_{0}^{9} x \, dx = \left[ 3 \cdot \frac{x^{3/2}}{3/2} \right]_{0}^{9} - \left[ \frac{x^2}{2} \right]_{0}^{9}$

$= \left[ 3 \cdot \frac{3^{3}}{3} \cdot 2 - 0 \right] - \left[ \frac{81}{2} - 0 \right]$

$= 54 - \frac{81}{2} = \frac{108 - 81}{2} = \frac{27}{2} \text{ sq. units}$