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

Options

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

Answer

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

### Step 1: Identify the curves
The given equations are:
1. The parabola: \(x^2 = 4y\)
2. The line: \(x = 4y - 2\)

### Step 2: Find the points of intersection
To find the points where the curves intersect, we can substitute \(y\) from the line equation into the parabola equation.

From the line equation:
\[ y = \frac{x + 2}{4} \]

Substituting this into the parabola equation:
\[ x^2 = 4\left(\frac{x + 2}{4}\right) \]
This simplifies to:
\[ x^2 = x + 2 \]
Rearranging gives:
\[ x^2 - x - 2 = 0 \]

### Step 3: Factor the quadratic equation
Factoring the quadratic:
\[ (x - 2)(x + 1) = 0 \]
Thus, the solutions are:
\[ x = 2 \quad \text{and} \quad x = -1 \]

### Step 4: Find corresponding \(y\) values
Now, we can find the \(y\) values for these \(x\) coordinates using the line equation:
1. For \(x = 2\):
\[ y = \frac{2 + 2}{4} = 1 \]
So, the point is \((2, 1)\).

2. For \(x = -1\):
\[ y = \frac{-1 + 2}{4} = \frac{1}{4} \]
So, the point is \((-1, \frac{1}{4})\).

### Step 5: Set up the integral for the area
The area \(A\) between the curves from \(x = -1\) to \(x = 2\) can be calculated by integrating the difference between the upper curve and the lower curve.

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

Thus, the area can be expressed as:
\[ A = \int_{-1}^{2} \left(\frac{x + 2}{4} - \frac{x^2}{4}\right) dx \]

### Step 6: Simplify and evaluate the integral
Factoring out \(\frac{1}{4}\):
\[ A = \frac{1}{4} \int_{-1}^{2} (x + 2 - x^2) dx \]

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

### Step 7: Calculate the definite integral
Calculating the upper limit (\(x = 2\)):
\[ \frac{2^2}{2} + 2(2) - \frac{2^3}{3} = 2 + 4 - \frac{8}{3} = 6 - \frac{8}{3} = \frac{18}{3} - \frac{8}{3} = \frac{10}{3} \]

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

### Step 8: Combine the results
Now, substituting back into the area formula:
\[ A = \frac{1}{4} \left( \frac{10}{3} - \left(-\frac{7}{6}\right) \right) = \frac{1}{4} \left( \frac{10}{3} + \frac{7}{6} \right) \]

Finding a common denominator (which is 6):
\[ \frac{10}{3} = \frac{20}{6} \]
Thus:
\[ A = \frac{1}{4} \left( \frac{20}{6} + \frac{7}{6} \right) = \frac{1}{4} \left( \frac{27}{6} \right) = \frac{27}{24} = \frac{9}{8} \]

### Final Answer
The area bounded by the curves is:
\[ \boxed{\frac{9}{8}} \text{ square units} \]