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: Set up the area idea.
The area trapped between two curves is the integral of the upper curve minus the lower curve, taken between the points where they cross. So first we find those crossing points.

Step 2: Find where they meet.
The line is $y = 2x$. Put this into the parabola $y^2 = 4x$.
\[ (2x)^2 = 4x \implies 4x^2 = 4x \implies 4x(x - 1) = 0 \] So $x = 0$ or $x = 1$. These are the limits.

Step 3: Decide which curve is on top.
Between $0$ and $1$, the parabola $y = 2\sqrt{x}$ sits above the line $y = 2x$. Test $x = \frac{1}{4}$: parabola gives $1$, line gives $\frac{1}{2}$. So parabola is higher.

Step 4: Write the area integral.
Subtract the lower curve from the upper curve and integrate.
\[ A = \int_0^1 (2\sqrt{x} - 2x)\,dx \]

Step 5: Integrate term by term.
The integral of $2\sqrt{x} = 2x^{1/2}$ is $\frac{4}{3}x^{3/2}$, and the integral of $2x$ is $x^2$.
\[ A = \left[\frac{4}{3}x^{3/2} - x^2\right]_0^1 \]

Step 6: Put in the limits.
At $x = 1$ we get $\frac{4}{3} - 1 = \frac{1}{3}$. At $x = 0$ everything is zero.
\[ A = \frac{1}{3} \]
\[ \boxed{\dfrac{1}{3}} \]