Find the area of the region bounded by the parabola `"x"^2=4"y\ "` and the line `"x"=4"y"-2`

Options

A. `9//8`
B. `9//4`
C. `9//2`
D. `9//7`

Answer

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

### Step 1: Identify the curves
The first curve is a parabola given by the equation:
\[ x^2 = 4y \]
This can be rewritten as:
\[ y = \frac{x^2}{4} \]

The second curve is a line given by the equation:
\[ x = 4y - 2 \]
This can be rearranged to find \( y \):
\[ y = \frac{x + 2}{4} \]

### Step 2: Find the points of intersection
To find the points where the parabola and the line intersect, we set the two equations for \( y \) equal to each other:
\[ \frac{x^2}{4} = \frac{x + 2}{4} \]
Multiplying through by 4 to eliminate the denominator:
\[ x^2 = x + 2 \]
Rearranging gives us:
\[ x^2 - x - 2 = 0 \]
Factoring the quadratic:
\[ (x - 2)(x + 1) = 0 \]
Thus, the solutions for \( x \) are:
\[ x = 2 \quad \text{and} \quad x = -1 \]

### Step 3: Find corresponding \( y \) values
Now we will find the corresponding \( y \) values for these \( x \) values using the parabola's equation \( y = \frac{x^2}{4} \):
- For \( x = 2 \):
\[ y = \frac{2^2}{4} = \frac{4}{4} = 1 \]
- For \( x = -1 \):
\[ y = \frac{(-1)^2}{4} = \frac{1}{4} \]

Thus, the points of intersection are:
\[ (-1, \frac{1}{4}) \quad \text{and} \quad (2, 1) \]

### Step 4: 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:
\[ A = \int_{-1}^{2} \left( \text{(upper function)} - \text{(lower function)} \right) \, dx \]
In this case, the upper function is the line \( y = \frac{x + 2}{4} \) and the lower function is the parabola \( y = \frac{x^2}{4} \):
\[ A = \int_{-1}^{2} \left( \frac{x + 2}{4} - \frac{x^2}{4} \right) \, dx \]
This simplifies to:
\[ A = \frac{1}{4} \int_{-1}^{2} (x + 2 - x^2) \, dx \]

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

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

### Step 6: Combine results
Now we combine the results:
\[ A = \frac{1}{4} \left( \frac{10}{3} + \frac{7}{24} \right) \]
Finding a common denominator:
\[ A = \frac{1}{4} \left( \frac{80}{24} + \frac{7}{24} \right) = \frac{1}{4} \left( \frac{87}{24} \right) = \frac{87}{96} \]

Thus, the area of the region bounded by the parabola and the line is:
\[ \boxed{\frac{87}{96}} \text{ square units} \]