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

$\frac{1}{2}$

The correct answer is Option (2) → $\frac{1}{2}$

We have, $y^2 = 9x \quad \text{and} \quad y = 3x$

On solving both the equations, we get

$⇒(3x)^2 = 9x$

$⇒9x^2 - 9x = 0$

$⇒9x(x - 1) = 0$

$⇒x = 0, 1$

Let $A_1 = \text{Area under the curve } y^2 = 9x$

and $A_2 = \text{Area under the line } y = 3x$

i.e., $A_1 = \text{Area of OABCO}$

and $A_2 = \text{Area of OBCO}$

$∴$ Required area $= A_1 - A_2$

$= \int\limits_{0}^{1} \sqrt{9x} \, dx - \int\limits_{0}^{1} 3x \, dx$

$= 3 \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{1} - 3 \left[ \frac{x^2}{2} \right]_{0}^{1} \qquad \left[ ∵\int x^n \, dx = \frac{x^{n+1}}{n+1} \right]$

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

$= 2 - \frac{3}{2} = \frac{1}{2} \text{ sq. unit}$