Question

The area of the region bounded by the curve $y^2 = 4x$ and the line $y = x$ is

• A. $\frac{8}{3}\text{ sq. units}$
• B. $\frac{5}{8}\text{ sq. units}$
• C. $\frac{3}{8}\text{ sq. units}$
• D. $\frac{3}{5}\text{ sq. units}$

Hint:

There is a direct shortcut formula for the area bounded by the parabola $y^2 = 4ax$ and the line $y = mx$: $\text{Area} = \frac{8a^2}{3m^3}$. Here, $4a = 4 \implies a = 1$, and $m = 1$. So, $\text{Area} = \frac{8(1)^2}{3(1)^3} = \frac{8}{3}$.

Answer 1

The Correct Option is A

Step 1: Find where they meet.
Put $y = x$ into $y^2 = 4x$: $x^2 = 4x$, so $x = 0$ or $x = 4$.

Step 2: Say which curve is on top.
Between 0 and 4 the parabola $y = 2\sqrt x$ is above the line $y = x$.

Step 3: Integrate the gap.
$$\text{Area} = \int_0^4 (2\sqrt x - x)\,dx = \left[\tfrac{4}{3}x^{3/2} - \tfrac{x^2}{2}\right]_0^4 = \frac{32}{3} - 8 = \frac{8}{3}$$
\[ \boxed{\dfrac{8}{3}\ \text{sq. units}} \]