The area of the region bounded by `y=x^(2)` and `y=-x^(2)+2` is

Options

A. `(5)/(3)`
B. `(4)/(3)`
C. `(8)/(3)`
D. 1

Answer

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

### Step 1: Find the Points of Intersection
To find the area between the two curves, we first need to determine where they intersect. We set the equations equal to each other:

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

Rearranging gives:

\[ x^2 + x^2 = 2 \implies 2x^2 = 2 \implies x^2 = 1 \implies x = \pm 1 \]

Thus, the points of intersection are \( x = 1 \) and \( x = -1 \). We can find the corresponding \( y \) values by substituting \( x = 1 \) into either equation:

\[ y = 1^2 = 1 \quad \text{or} \quad y = -1^2 + 2 = 1 \]

So, the points of intersection are \( (1, 1) \) and \( (-1, 1) \).

### Step 2: Set Up the Integral for Area
The area \( A \) between the curves from \( x = -1 \) to \( x = 1 \) can be calculated using the integral:

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

This simplifies to:

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

### Step 3: Calculate the Integral
Now we will compute the integral:

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

This can be split into two separate integrals:

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

Calculating each integral separately:

1. **For the first integral:**
\[ \int -2x^2 \, dx = -\frac{2}{3} x^3 \Big|_{-1}^{1} = -\frac{2}{3}(1^3 - (-1)^3) = -\frac{2}{3}(1 - (-1)) = -\frac{2}{3}(2) = -\frac{4}{3} \]

2. **For the second integral:**
\[ \int 2 \, dx = 2x \Big|_{-1}^{1} = 2(1 - (-1)) = 2(2) = 4 \]

Now, combining these results:

\[ A = -\frac{4}{3} + 4 = -\frac{4}{3} + \frac{12}{3} = \frac{8}{3} \]

### Step 4: Final Area Calculation
Since the area is always positive, we take the absolute value:

\[ A = \frac{8}{3} \]

Thus, the area of the region bounded by the curves \( y = x^2 \) and \( y = -x^2 + 2 \) is:

\[ \boxed{\frac{8}{3}} \text{ square units} \]