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 will follow these steps:

### Step 1: Find the points of intersection
We start by finding the points where the curve and the line intersect.

1. The equation of the parabola is given by:
\[ x^2 = 4y \]
We can express \( y \) in terms of \( x \):
\[ y = \frac{x^2}{4} \]

2. The equation of the line is:
\[ x = 4y - 2 \implies 4y = x + 2 \implies y = \frac{x + 2}{4} \]

3. Set the two expressions for \( y \) equal to find the intersection points:
\[ \frac{x^2}{4} = \frac{x + 2}{4} \]

4. Multiply through by 4 to eliminate the denominator:
\[ x^2 = x + 2 \]

5. Rearranging gives us:
\[ x^2 - x - 2 = 0 \]

6. Factor the quadratic:
\[ (x - 2)(x + 1) = 0 \]

7. Thus, the solutions for \( x \) are:
\[ x = 2 \quad \text{and} \quad x = -1 \]

### Step 2: Find the corresponding \( y \) values
Now, we find the \( y \) values corresponding to these \( x \) values.

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 calculated using the formula:
\[ A = \int_{-1}^{2} \left( \text{upper curve} - \text{lower curve} \right) \, dx \]

1. The upper curve is the line \( y = \frac{x + 2}{4} \) and the lower curve is the parabola \( y = \frac{x^2}{4} \).

2. Thus, we have:
\[ A = \int_{-1}^{2} \left( \frac{x + 2}{4} - \frac{x^2}{4} \right) \, dx \]

3. Simplifying the integrand:
\[ A = \int_{-1}^{2} \frac{1}{4} (x + 2 - x^2) \, dx = \frac{1}{4} \int_{-1}^{2} (x + 2 - x^2) \, dx \]

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

1. Calculate the upper limit \( x = 2 \):
\[ \frac{2^2}{2} + 2(2) - \frac{2^3}{3} = 2 + 4 - \frac{8}{3} = 6 - \frac{8}{3} = \frac{18}{3} - \frac{8}{3} = \frac{10}{3} \]

2. Calculate the lower limit \( x = -1 \):
\[ \frac{(-1)^2}{2} + 2(-1) - \frac{(-1)^3}{3} = \frac{1}{2} - 2 + \frac{1}{3} = \frac{1}{2} - \frac{6}{3} + \frac{1}{3} = \frac{1}{2} - \frac{5}{3} = \frac{3}{6} - \frac{10}{6} = -\frac{7}{6} \]

3. Now substituting back into the integral:
\[ A = \frac{1}{4} \left( \frac{10}{3} - \left( -\frac{7}{6} \right) \right) = \frac{1}{4} \left( \frac{10}{3} + \frac{7}{6} \right) \]

4. Find a common denominator (which is 6):
\[ = \frac{1}{4} \left( \frac{20}{6} + \frac{7}{6} \right) = \frac{1}{4} \left( \frac{27}{6} \right) = \frac{27}{24} = \frac{9}{8} \]

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