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: See the two curves.
We need the area between the parabola $y^2=4x$ and the line $y=x$.
Step 2: Find where they meet, in $y$.
From the parabola $x=\frac{y^2}{4}$, and from the line $x=y$. Setting equal: $\frac{y^2}{4}=y$, so $y^2=4y$, giving $y=0$ and $y=4$.
Step 3: Choose to integrate along $y$.
For each $y$ from $0$ to $4$, the line $x=y$ is on the right and the parabola $x=\frac{y^2}{4}$ is on the left. So width $=y-\frac{y^2}{4}$.
Step 4: Set up the integral.
Area $=\displaystyle\int_0^4\left(y-\frac{y^2}{4}\right)dy$.
Step 5: Integrate.
$=\left[\dfrac{y^2}{2}-\dfrac{y^3}{12}\right]_0^4=\dfrac{16}{2}-\dfrac{64}{12}=8-\dfrac{16}{3}$.
Step 6: Final value.
$8-\dfrac{16}{3}=\dfrac{24-16}{3}=\dfrac{8}{3}$ sq. units, which is option (1). \[ \boxed{\text{Area}=\frac{8}{3}\text{ sq. units}} \]