The area bounded by the curve $y^{2}=x^{3}$ and the line $y=2x$ is :

Question

Accepted Answer

To determine the area bounded by the curve $ y^2 = x^3 $ and the line $ y = 2x $, we need to find the points of intersection and then calculate the area between these curves.

### Step 1: Find Points of Intersection

1. **Equation of the curve:** $ y^2 = x^3 $
2. **Equation of the line:** $ y = 2x $

Substitute $ y = 2x $ into the curve equation:

$$
(2x)^2 = x^3 \implies 4x^2 = x^3
$$

Rearrange and factor:

$$
x^3 - 4x^2 = 0 \implies x^2(x - 4) = 0
$$

This gives the solutions $ x = 0 $ and $ x = 4 $.

For $ x = 0 $, $ y = 2 \times 0 = 0 $. So, one point of intersection is $ (0, 0) $.

For $ x = 4 $, $ y = 2 \times 4 = 8 $. So, the other point of intersection is $ (4, 8) $.

### Step 2: Calculate the Area

The area between the curves from $ x = 0 $ to $ x = 4 $ is given by the integral of the difference between the line and the curve:

$$
\text{Area} = \int_{0}^{4} (2x - \sqrt{x^3}) \, dx
$$

Calculate each integral separately:

1. **Integral of the line:**

$$
\int 2x \, dx = x^2
$$

2. **Integral of the curve:**

$$
\int \sqrt{x^3} \, dx = \int x^{3/2} \, dx = \frac{2}{5}x^{5/2}
$$

Evaluate the definite integrals from 0 to 4:

$$
\text{Area} = \left[ x^2 \right]_{0}^{4} - \left[ \frac{2}{5}x^{5/2} \right]_{0}^{4}
$$

Calculate each part:

- For $ x^2 $:

$$
\left[ x^2 \right]_{0}^{4} = 4^2 - 0^2 = 16
$$

- For $ \frac{2}{5}x^{5/2} $:

$$
\left[ \frac{2}{5}x^{5/2} \right]_{0}^{4} = \frac{2}{5}(4^{5/2}) - \frac{2}{5}(0^{5/2}) = \frac{2}{5}(32) = \frac{64}{5}
$$

Subtract the results:

$$
\text{Area} = 16 - \frac{64}{5} = \frac{80}{5} - \frac{64}{5} = \frac{16}{5}
$$

Thus, the area bounded by the curve $ y^2 = x^3 $ and the line $ y = 2x $ is $\frac{16}{5}$.

### Conclusion

The correct answer is Option 2: $\frac{16}{5}$.

### Explanation of Incorrect Options

- **Option 1: $\frac{16}{3}$** is incorrect because the calculation of the area does not yield this value.
- **Option 3: $\frac{8}{3}$** is incorrect as it does not match the calculated area.
- **Option 4: $\frac{8}{5}$** is incorrect because the area calculation results in $\frac{16}{5}$, not $\frac{8}{5}$.

Therefore, the correct answer is indeed Option 2: $\frac{16}{5}$.