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

Options

A. `3/8"sq units"`
B. `5/8" sq unit "`
C. `7/8 " sq unit"`
D. `9/8" sq units"`

Explanation

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

### Step 1: Find the points of intersection
To find the points where the curve and the line intersect, we need to solve the equations simultaneously.

1. From the curve \(x^2 = 4y\), we can express \(y\) in terms of \(x\):
\[ y = \frac{x^2}{4} \]

2. Substitute this expression for \(y\) into the line equation \(x = 4y - 2\):
\[ x = 4\left(\frac{x^2}{4}\right) - 2 \]
Simplifying this gives:
\[ x = x^2 - 2 \]
Rearranging leads to:
\[ x^2 - x - 2 = 0 \]

3. Now, we can factor the quadratic:
\[ (x - 2)(x + 1) = 0 \]
Thus, the solutions are:
\[ x = 2 \quad \text{and} \quad x = -1 \]

### Step 2: Find the corresponding \(y\) values
Now we need to find the \(y\) values corresponding to these \(x\) values using the equation \(y = \frac{x^2}{4}\).

1. For \(x = 2\):
\[ y = \frac{2^2}{4} = \frac{4}{4} = 1 \]

2. For \(x = -1\):
\[ y = \frac{(-1)^2}{4} = \frac{1}{4} \]

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

### Step 3: Set up the integral for the area
The area \(A\) between the curves from \(x = -1\) to \(x = 2\) can be found using the integral:
\[ A = \int_{-1}^{2} (y_{\text{line}} - y_{\text{curve}}) \, dx \]
Where \(y_{\text{line}} = \frac{x + 2}{4}\) (from rearranging \(x = 4y - 2\)) and \(y_{\text{curve}} = \frac{x^2}{4}\).

### Step 4: Evaluate the integral
Now we can set up the integral:
\[ A = \int_{-1}^{2} \left(\frac{x + 2}{4} - \frac{x^2}{4}\right) \, dx \]
This simplifies to:
\[ A = \frac{1}{4} \int_{-1}^{2} (x + 2 - x^2) \, dx \]

### Step 5: Calculate the integral
Now we calculate the integral:
\[ A = \frac{1}{4} \left[ \frac{x^2}{2} + 2x - \frac{x^3}{3} \right]_{-1}^{2} \]

Calculating at the upper limit \(x = 2\):
\[ = \frac{1}{4} \left[ \frac{2^2}{2} + 2(2) - \frac{2^3}{3} \right] = \frac{1}{4} \left[ 2 + 4 - \frac{8}{3} \right] = \frac{1}{4} \left[ 6 - \frac{8}{3} \right] = \frac{1}{4} \left[ \frac{18}{3} - \frac{8}{3} \right] = \frac{1}{4} \left[ \frac{10}{3} \right] = \frac{10}{12} = \frac{5}{6} \]

Calculating at the lower limit \(x = -1\):
\[ = \frac{1}{4} \left[ \frac{(-1)^2}{2} + 2(-1) - \frac{(-1)^3}{3} \right] = \frac{1}{4} \left[ \frac{1}{2} - 2 + \frac{1}{3} \right] \]
Finding a common denominator (6):
\[ = \frac{1}{4} \left[ \frac{3}{6} - \frac{12}{6} + \frac{2}{6} \right] = \frac{1}{4} \left[ \frac{-7}{6} \right] = -\frac{7}{24} \]

### Step 6: Combine results
Now we combine the results:
\[ A = \frac{5}{6} - \left(-\frac{7}{24}\right) = \frac{5}{6} + \frac{7}{24} \]
Finding a common denominator (24):
\[ = \frac{20}{24} + \frac{7}{24} = \frac{27}{24} \]

### Final Result
Thus, the area of the region bounded by the curve and the line is:
\[ \frac{27}{24} \text{ square units} \]