The area of the region bounded by the curve `x^(2)=4y` and the straight line `x=4y-2` is

Options

A. `5/4`
B. `9/8`
C. `3/4`
D. `7/8`

Answer

To find the area of the region bounded by the curve \( x^2 = 4y \) and the straight line \( x = 4y - 2 \), we will follow these steps:

### Step 1: Find the points of intersection
To find the area between the two curves, we first need to determine where they intersect.

1. **Substitute \( y \) from the curve equation into the line equation**:
- From \( x^2 = 4y \), we can express \( y \) as:
\[ y = \frac{x^2}{4} \]
- Substitute this into the line equation \( x = 4y - 2 \):
\[ x = 4\left(\frac{x^2}{4}\right) - 2 \]
\[ x = x^2 - 2 \]
\[ x^2 - x - 2 = 0 \]

2. **Solve the quadratic equation**:
\[ (x - 2)(x + 1) = 0 \]
Thus, \( x = 2 \) and \( x = -1 \).

3. **Find corresponding \( y \) values**:
- For \( x = 2 \):
\[ y = \frac{2^2}{4} = 1 \]
- For \( x = -1 \):
\[ y = \frac{(-1)^2}{4} = \frac{1}{4} \]

So, the points of intersection are \( A(-1, \frac{1}{4}) \) and \( B(2, 1) \).

### Step 2: Set up the integral for the area
The area \( A \) between the curves from \( x = -1 \) to \( x = 2 \) can be found by integrating the difference of the functions:

1. **Identify the upper and lower functions**:
- The line \( y = \frac{x + 2}{4} \) is above the parabola \( y = \frac{x^2}{4} \) in the interval from \( x = -1 \) to \( x = 2 \).

2. **Set up the integral**:
\[ A = \int_{-1}^{2} \left(\frac{x + 2}{4} - \frac{x^2}{4}\right) \, dx \]
Simplifying the integrand:
\[ A = \int_{-1}^{2} \frac{x + 2 - x^2}{4} \, dx = \frac{1}{4} \int_{-1}^{2} (2 + x - x^2) \, dx \]

### Step 3: Evaluate the integral
1. **Integrate**:
\[ \int (2 + x - x^2) \, dx = 2x + \frac{x^2}{2} - \frac{x^3}{3} \]

2. **Evaluate from \( -1 \) to \( 2 \)**:
\[ A = \frac{1}{4} \left[ \left(2(2) + \frac{2^2}{2} - \frac{2^3}{3}\right) - \left(2(-1) + \frac{(-1)^2}{2} - \frac{(-1)^3}{3}\right) \right] \]
Calculating the first part:
\[ = \left(4 + 2 - \frac{8}{3}\right) = 6 - \frac{8}{3} = \frac{18 - 8}{3} = \frac{10}{3} \]
Calculating the second part:
\[ = \left(-2 + \frac{1}{2} + \frac{1}{3}\right) = -2 + \frac{3}{6} + \frac{2}{6} = -2 + \frac{5}{6} = -\frac{12}{6} + \frac{5}{6} = -\frac{7}{6} \]

3. **Combine the results**:
\[ A = \frac{1}{4} \left(\frac{10}{3} + \frac{7}{6}\right) \]
Finding a common denominator:
\[ = \frac{1}{4} \left(\frac{20}{6} + \frac{7}{6}\right) = \frac{1}{4} \left(\frac{27}{6}\right) = \frac{27}{24} = \frac{9}{8} \]

### Final Answer
The area of the region bounded by the curve \( x^2 = 4y \) and the line \( x = 4y - 2 \) is \( \frac{9}{8} \) square units.

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