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: Understanding the Question:
We must calculate the enclosed area situated between a rightward-opening parabola ($y^2 = 4x$) and a straight line passing through the origin ($y = x$).

Step 2: Key Formula or Approach:
First, find the points of intersection by setting the equations equal to each other.
Then, set up a definite integral to find the area between the curves:
$$\text{Area} = \int_a^b (\text{Upper Curve} - \text{Lower Curve}) \, dx$$ For the parabola $y^2 = 4x$ in the first quadrant, the upper branch is $y = \sqrt{4x} = 2\sqrt{x}$.

Step 3: Detailed Explanation:
Find intersection points by substituting $y = x$ into $y^2 = 4x$:
$$x^2 = 4x$$ $$x^2 - 4x = 0$$ $$x(x - 4) = 0$$ Thus, the curves intersect at $x = 0$ and $x = 4$.
Between $x=0$ and $x=4$, test a point (e.g., $x=1$):
For the parabola, $y = 2\sqrt{1} = 2$.
For the line, $y = 1$.
Since $2 > 1$, the parabola is the upper curve and the line is the lower curve.
Set up the area integral:
$$\text{Area} = \int_0^4 (2\sqrt{x} - x) \, dx$$ Integrate term by term:
$$\text{Area} = \left[ 2 \left(\frac{x^{3/2}}{3/2}\right) - \frac{x^2}{2} \right]_0^4$$ $$\text{Area} = \left[ \frac{4}{3}x^{3/2} - \frac{x^2}{2} \right]_0^4$$ Substitute the upper limit ($x=4$):
$$\text{Area} = \left( \frac{4}{3}(4)^{3/2} - \frac{4^2}{2} \right) - (0)$$ Since $4^{3/2} = (\sqrt{4})^3 = 2^3 = 8$:
$$\text{Area} = \left( \frac{4}{3}(8) - \frac{16}{2} \right)$$ $$\text{Area} = \frac{32}{3} - 8$$ Get a common denominator:
$$\text{Area} = \frac{32}{3} - \frac{24}{3} = \frac{8}{3}$$

Step 4: Final Answer:
The area of the bounded region is $\frac{8}{3}\text{ sq. units}$, corresponding to option (A).