The area bounded by `y=x^(2)+2 and y=2|x|-cospi x` is equal to

Options

A. `2//3`
B. `8//3`
C. `4//3`
D. `1//3`

Explanation

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

### Step 1: Find the points of intersection
We need to set the two equations equal to each other to find the points of intersection:
\[ x^2 + 2 = 2|x| - \cos(\pi x) \]

### Step 2: Analyze the equation for different cases
Since \( |x| \) behaves differently for positive and negative values of \( x \), we will consider two cases:

**Case 1:** \( x \geq 0 \) (thus \( |x| = x \))
\[ x^2 + 2 = 2x - \cos(\pi x) \]
Rearranging gives:
\[ x^2 - 2x + 2 + \cos(\pi x) = 0 \]

**Case 2:** \( x < 0 \) (thus \( |x| = -x \))
\[ x^2 + 2 = -2x - \cos(\pi x) \]
Rearranging gives:
\[ x^2 + 2x + 2 + \cos(\pi x) = 0 \]

### Step 3: Solve for intersection points
We will solve these equations to find the intersection points.

For both cases, we can observe that \( \cos(\pi x) \) oscillates between -1 and 1. Thus, we can find approximate solutions:
- For \( x = -1 \):
\[ (-1)^2 + 2 = 1 + 2 = 3 \]
\[ 2|-1| - \cos(-\pi) = 2 + 1 = 3 \quad \text{(valid)} \]

- For \( x = 1 \):
\[ (1)^2 + 2 = 1 + 2 = 3 \]
\[ 2|1| - \cos(\pi) = 2 + 1 = 3 \quad \text{(valid)} \]

Thus, the points of intersection are \( x = -1 \) and \( x = 1 \).

### Step 4: Set up the integral for the area
The area \( A \) between the curves from \( x = -1 \) to \( x = 1 \) is given by:
\[ A = \int_{-1}^{1} \left( (2|x| - \cos(\pi x)) - (x^2 + 2) \right) \, dx \]

### Step 5: Simplify the integral
Substituting \( |x| \) gives:
\[ A = \int_{-1}^{1} \left( (2x - \cos(\pi x)) - (x^2 + 2) \right) \, dx \quad \text{for } x \geq 0 \]
\[ = \int_{0}^{1} \left( (2x - \cos(\pi x)) - (x^2 + 2) \right) \, dx + \int_{-1}^{0} \left( (-2x - \cos(\pi x)) - (x^2 + 2) \right) \, dx \]

### Step 6: Calculate the integral
Evaluating the integral:
\[ A = 2 \int_{0}^{1} \left( 2x - x^2 - 2 - \cos(\pi x) \right) \, dx \]

Calculating this integral:
1. The integral of \( 2x \) from 0 to 1 is \( [x^2]_{0}^{1} = 1 \).
2. The integral of \( x^2 \) from 0 to 1 is \( [\frac{x^3}{3}]_{0}^{1} = \frac{1}{3} \).
3. The integral of \( -2 \) from 0 to 1 is \( [-2x]_{0}^{1} = -2 \).
4. The integral of \( -\cos(\pi x) \) from 0 to 1 is \( [-\frac{1}{\pi} \sin(\pi x)]_{0}^{1} = 0 \).

Putting it all together:
\[ A = 2 \left( 1 - \frac{1}{3} - 2 \right) = 2 \left( -\frac{4}{3} \right) = -\frac{8}{3} \]
Since we are interested in the area, we take the absolute value:
\[ A = \frac{8}{3} \]

### Final Answer
The area bounded by the curves is \( \frac{8}{3} \).