Area bounded by $x^{2}=4y$ and $x=4y-2$ is:

Question

Accepted Answer

To determine the area bounded by the curves $x^2 = 4y$ and $x = 4y - 2$, 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 Parabola:** $x^2 = 4y$ can be rewritten as $y = \frac{x^2}{4}$.
2. **Equation of the Line:** $x = 4y - 2$ can be rewritten as $y = \frac{x + 2}{4}$.

To find the points of intersection, set the equations for $y$ equal to each other:

$$
\frac{x^2}{4} = \frac{x + 2}{4}
$$

Multiply through by 4 to clear the denominators:

$$
x^2 = x + 2
$$

Rearrange to form a quadratic equation:

$$
x^2 - x - 2 = 0
$$

Factor the quadratic:

$$
(x - 2)(x + 1) = 0
$$

Thus, the solutions are $x = 2$ and $x = -1$.

### Step 2: Calculate the Area

The area between the curves from $x = -1$ to $x = 2$ is given by the integral of the difference between the top function and the bottom function. Here, the line $y = \frac{x + 2}{4}$ is above the parabola $y = \frac{x^2}{4}$ in this interval.

The area $A$ is:

$$
A = \int_{-1}^{2} \left(\frac{x + 2}{4} - \frac{x^2}{4}\right) \, dx
$$

Simplify the integrand:

$$
A = \int_{-1}^{2} \left(\frac{x + 2 - x^2}{4}\right) \, dx = \frac{1}{4} \int_{-1}^{2} (x + 2 - x^2) \, dx
$$

Calculate the integral:

$$
A = \frac{1}{4} \left[ \frac{x^2}{2} + 2x - \frac{x^3}{3} \right]_{-1}^{2}
$$

Evaluate the definite integral:

1. At $x = 2$:

$$
\frac{2^2}{2} + 2(2) - \frac{2^3}{3} = 2 + 4 - \frac{8}{3} = 6 - \frac{8}{3} = \frac{18}{3} - \frac{8}{3} = \frac{10}{3}
$$

2. At $x = -1$:

$$
\frac{(-1)^2}{2} + 2(-1) - \frac{(-1)^3}{3} = \frac{1}{2} - 2 + \frac{1}{3} = \frac{1}{2} - 2 + \frac{1}{3} = \frac{3}{6} - \frac{12}{6} + \frac{2}{6} = -\frac{7}{6}
$$

Subtract the results:

$$
\frac{10}{3} - \left(-\frac{7}{6}\right) = \frac{10}{3} + \frac{7}{6} = \frac{20}{6} + \frac{7}{6} = \frac{27}{6} = \frac{9}{2}
$$

Multiply by $\frac{1}{4}$:

$$
A = \frac{1}{4} \times \frac{9}{2} = \frac{9}{8}
$$

Thus, the area bounded by the curves is $\frac{9}{8}$, which corresponds to Option 3.

### Explanation of Incorrect Options

- **Option 1: $\frac{1}{8}$**: This value is too small compared to the calculated area.
- **Option 2: $\frac{8}{9}$**: This value is incorrect as it does not match the calculated area.
- **Option 4: $\frac{1}{9}$**: This value is also too small and does not match the calculated area.

Therefore, the correct answer is indeed Option 3: $\frac{9}{8}$.