The area (in square units) of the region bounded by the curves `y+2x^(2)=0, y+3x^(2)=1` is equal to

Options

A. `(3)/(5)`
B. `(3)/(4)`
C. `(1)/(3)`
D. `(4)/(3)`

Answer

To find the area of the region bounded by the curves \( y + 2x^2 = 0 \) and \( y + 3x^2 = 1 \), we will follow these steps:

### Step 1: Rewrite the equations of the curves
First, we rewrite the equations of the curves in terms of \( y \):
1. From \( y + 2x^2 = 0 \), we get:
\[ y = -2x^2 \]
2. From \( y + 3x^2 = 1 \), we get:
\[ y = 1 - 3x^2 \]

### Step 2: Find the points of intersection
To find the points of intersection, we set the two equations equal to each other:
\[ -2x^2 = 1 - 3x^2 \]
Rearranging gives:
\[ 3x^2 - 2x^2 = 1 \implies x^2 = 1 \implies x = \pm 1 \]
Now, substituting \( x = 1 \) into either equation to find \( y \):
\[ y = -2(1)^2 = -2 \quad \text{and} \quad y = 1 - 3(1)^2 = -2 \]
Thus, the points of intersection are \( (1, -2) \) and \( (-1, -2) \).

### Step 3: Set up the integral for the area
The area \( A \) between the curves from \( x = -1 \) to \( x = 1 \) can be calculated using the integral:
\[ A = \int_{-1}^{1} \left( (1 - 3x^2) - (-2x^2) \right) \, dx \]
This simplifies to:
\[ A = \int_{-1}^{1} (1 - 3x^2 + 2x^2) \, dx = \int_{-1}^{1} (1 - x^2) \, dx \]

### Step 4: Evaluate the integral
Now we evaluate the integral:
\[ A = \int_{-1}^{1} (1 - x^2) \, dx \]
Calculating the integral:
\[ = \left[ x - \frac{x^3}{3} \right]_{-1}^{1} \]
Calculating at the boundaries:
\[ = \left( 1 - \frac{1^3}{3} \right) - \left( -1 + \frac{(-1)^3}{3} \right) \]
\[ = \left( 1 - \frac{1}{3} \right) - \left( -1 - \frac{1}{3} \right) \]
\[ = \left( \frac{2}{3} \right) - \left( -\frac{4}{3} \right) = \frac{2}{3} + \frac{4}{3} = \frac{6}{3} = 2 \]

### Step 5: Calculate the total area
Since the area is symmetric about the y-axis, we can multiply the area calculated from \( 0 \) to \( 1 \) by \( 2 \):
\[ \text{Total Area} = 2 \]

Thus, the area of the region bounded by the curves is \( \boxed{2} \).