Question

The area (in square units) of the region bounded by the parabola $y^2 = 4x$ and the line $y = 2x$ is:

• A. $\frac{1}{3}$
• B. $\frac{1}{6}$
• C. $\frac{2}{3}$
• D. $\frac{4}{3}$

Hint:

The area bounded by the parabola $y^2 = 4ax$ and the line $y = mx$ is given by the direct formula $\frac{8a^2}{3m^3}$. Here, $4a = 4 \implies a = 1$ and $m = 2$, so $\frac{8(1)^2}{3(2)^3} = \frac{8}{24} = \frac{1}{3}$.

Answer 1

The Correct Option is A

Step 1: Concept
The area $A$ of the region bounded by two curves $y = f(x)$ and $y = g(x)$ between their intersection points $x = a$ and $x = b$ is given by: \[ A = \int_a^b |f(x) - g(x)| \, dx \]

Step 2: Meaning
We first find the points of intersection of the parabola $y^2 = 4x \implies y = 2\sqrt{x}$ (upper curve for $x \ge 0$) and the line $y = 2x$ (lower curve).

Step 3: Analysis
Finding the intersection points: \[ (2x)^2 = 4x \implies 4x^2 = 4x \implies 4x(x - 1) = 0 \implies x = 0 \text{ and } x = 1 \] Thus, the limits of integration are from $0$ to $1$. Calculating the area: \[ A = \int_0^1 (2\sqrt{x} - 2x) \, dx \] \[ A = \left[ 2 \cdot \frac{x^{3/2}}{3/2} - x^2 \right]_0^1 \] \[ A = \left[ \frac{4}{3}x^{3/2} - x^2 \right]_0^1 = \left( \frac{4}{3} - 1 \right) - (0) = \frac{1}{3} \]

Step 4: Conclusion
The area of the bounded region is $\frac{1}{3}$ square units. Final Answer: (A)