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: Identify the curves
The first curve is a parabola given by \( x^2 = 4y \). This can be rewritten as \( y = \frac{x^2}{4} \).

The second curve is a line given by \( x = 4y - 2 \). Rearranging gives \( y = \frac{x + 2}{4} \).

### Step 2: Find the points of intersection
To find the area between the curves, we need to find their points of intersection. We set the equations equal to each other:

\[ \frac{x^2}{4} = \frac{x + 2}{4} \]

Multiplying through by 4 to eliminate the denominators:

\[ x^2 = x + 2 \]

Rearranging gives:

\[ x^2 - x - 2 = 0 \]

Factoring the quadratic:

\[ (x - 2)(x + 1) = 0 \]

Thus, the solutions are:

\[ x = 2 \quad \text{and} \quad x = -1 \]

### Step 3: Determine the corresponding y-values
Now we substitute these x-values back into either equation to find the 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 \( (2, 1) \) and \( (-1, \frac{1}{4}) \).

### Step 4: Set up the integral for the area
The area \( A \) between the curves from \( x = -1 \) to \( x = 2 \) can be expressed as:

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

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

Thus, we have:

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

### Step 5: Simplify the integral
This simplifies to:

\[ A = \int_{-1}^{2} \frac{x + 2 - x^2}{4} \, dx = \frac{1}{4} \int_{-1}^{2} (2 + x - x^2) \, dx \]

### Step 6: Compute the integral
Now we compute the integral:

\[ \int (2 + x - x^2) \, dx = 2x + \frac{x^2}{2} - \frac{x^3}{3} \]

Evaluating from \( -1 \) to \( 2 \):

\[ = \left[ 2(2) + \frac{2^2}{2} - \frac{2^3}{3} \right] - \left[ 2(-1) + \frac{(-1)^2}{2} - \frac{(-1)^3}{3} \right] \]

Calculating the upper limit:

\[ = 4 + 2 - \frac{8}{3} = 6 - \frac{8}{3} = \frac{18}{3} - \frac{8}{3} = \frac{10}{3} \]

Calculating the lower limit:

\[ = -4 + \frac{1}{2} + \frac{1}{3} = -4 + \frac{3}{6} + \frac{2}{6} = -4 + \frac{5}{6} = -\frac{24}{6} + \frac{5}{6} = -\frac{19}{6} \]

Putting it all together:

\[ A = \frac{1}{4} \left( \frac{10}{3} + \frac{19}{6} \right) \]

Finding a common denominator (which is 6):

\[ = \frac{1}{4} \left( \frac{20}{6} + \frac{19}{6} \right) = \frac{1}{4} \left( \frac{39}{6} \right) = \frac{39}{24} = \frac{13}{8} \]

### Final Answer
The area of the region bounded by the curve \( x^2 = 4y \) and the line \( x = 4y - 2 \) is:

\[ \boxed{\frac{13}{8}} \text{ square units} \]