The area bounded by the curve `x^(2)=4y+4` and line `3x+4y=0` is

Options

A. `(25)/(4)" sq. units"`
B. `(125)/(8)" sq. units"`
C. `(125)/(16)" sq. units"`
D. `(125)/(24)" sq. units"`

Answer

To find the area bounded by the curve \( x^2 = 4y + 4 \) and the line \( 3x + 4y = 0 \), we will follow these steps:

### Step 1: Rewrite the equations
First, we rewrite the equations in a more usable form.

1. The parabola:
\[ x^2 = 4y + 4 \implies y = \frac{x^2}{4} - 1 \]

2. The line:
\[ 3x + 4y = 0 \implies y = -\frac{3}{4}x \]

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

\[ \frac{x^2}{4} - 1 = -\frac{3}{4}x \]

Multiplying through by 4 to eliminate the fraction:

\[ x^2 - 4 = -3x \]
\[ x^2 + 3x - 4 = 0 \]

Now, we can factor or use the quadratic formula to find the roots:

\[ (x + 4)(x - 1) = 0 \]

Thus, the solutions are:

\[ x = -4 \quad \text{and} \quad x = 1 \]

### Step 3: Set up the integral for the area
The area \( A \) between the curves from \( x = -4 \) to \( x = 1 \) can be calculated using the integral:

\[ A = \int_{-4}^{1} \left( \text{Top function} - \text{Bottom function} \right) \, dx \]

Here, the top function is the parabola and the bottom function is the line:

\[ A = \int_{-4}^{1} \left( \left( \frac{x^2}{4} - 1 \right) - \left( -\frac{3}{4}x \right) \right) \, dx \]

This simplifies to:

\[ A = \int_{-4}^{1} \left( \frac{x^2}{4} + \frac{3}{4}x - 1 \right) \, dx \]

### Step 4: Compute the integral
Now we compute the integral:

\[ A = \int_{-4}^{1} \left( \frac{x^2}{4} + \frac{3}{4}x - 1 \right) \, dx \]

Calculating the antiderivative:

\[ = \left[ \frac{x^3}{12} + \frac{3x^2}{8} - x \right]_{-4}^{1} \]

Now, we evaluate it at the bounds:

1. At \( x = 1 \):
\[ \frac{1^3}{12} + \frac{3(1^2)}{8} - 1 = \frac{1}{12} + \frac{3}{8} - 1 = \frac{1}{12} + \frac{9}{24} - \frac{24}{24} = \frac{1}{12} - \frac{15}{24} = \frac{1}{12} - \frac{5}{8} = \frac{1}{12} - \frac{15}{24} = \frac{2 - 15}{24} = \frac{-13}{24} \]

2. At \( x = -4 \):
\[ \frac{(-4)^3}{12} + \frac{3(-4)^2}{8} - (-4) = \frac{-64}{12} + \frac{48}{8} + 4 = \frac{-64}{12} + 6 + 4 = \frac{-64}{12} + 10 = \frac{-64 + 120}{12} = \frac{56}{12} = \frac{14}{3} \]

Now, substituting back into the integral:

\[ A = \left( \frac{-13}{24} - \frac{14}{3} \right) \]

Convert \( \frac{14}{3} \) to a fraction with a denominator of 24:

\[ \frac{14}{3} = \frac{112}{24} \]

Thus,

\[ A = \frac{-13}{24} - \frac{112}{24} = \frac{-125}{24} \]

Since area cannot be negative, we take the absolute value:

\[ A = \frac{125}{24} \]

### Final Answer
The area bounded by the curve \( x^2 = 4y + 4 \) and the line \( 3x + 4y = 0 \) is:

\[ \boxed{\frac{125}{24}} \text{ square units} \]