Using integration , find the area bounded between the curve ` y = x^(2) and y=- |x| + 2 `

Answer

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

### Step 1: Identify the curves
The first curve is a parabola given by \( y = x^2 \), which opens upwards. The second curve is a V-shaped line represented by \( y = -|x| + 2 \). This can be expressed as:
- For \( x \geq 0 \): \( y = -x + 2 \)
- For \( x < 0 \): \( y = x + 2 \)

### Step 2: Find the points of intersection
To find the area between the curves, we need to determine where they intersect. This requires solving the equations:
1. For \( x \geq 0 \): Set \( x^2 = -x + 2 \)
\[ x^2 + x - 2 = 0 \]
Factoring gives:
\[ (x - 1)(x + 2) = 0 \]
Thus, \( x = 1 \) (valid for \( x \geq 0 \)) and \( x = -2 \) (not valid for \( x \geq 0 \)).

2. For \( x < 0 \): Set \( x^2 = x + 2 \)
\[ x^2 - x - 2 = 0 \]
Factoring gives:
\[ (x - 2)(x + 1) = 0 \]
Thus, \( x = 2 \) (not valid for \( x < 0 \)) and \( x = -1 \) (valid for \( x < 0 \)).

The points of intersection are \( x = 1 \) and \( x = -1 \).

### Step 3: Set up the integral for area calculation
The area \( A \) between the curves from \( x = -1 \) to \( x = 1 \) can be calculated by integrating the difference of the upper curve and the lower curve.

From \( x = -1 \) to \( x = 0 \):
- Upper curve: \( y = x + 2 \)
- Lower curve: \( y = x^2 \)

From \( x = 0 \) to \( x = 1 \):
- Upper curve: \( y = -x + 2 \)
- Lower curve: \( y = x^2 \)

Thus, the area \( A \) can be expressed as:
\[ A = \int_{-1}^{0} ((x + 2) - (x^2)) \, dx + \int_{0}^{1} ((-x + 2) - (x^2)) \, dx \]

### Step 4: Calculate the first integral
\[ \int_{-1}^{0} (x + 2 - x^2) \, dx = \int_{-1}^{0} (-x^2 + x + 2) \, dx \]
Calculating this integral:
\[ = \left[ -\frac{x^3}{3} + \frac{x^2}{2} + 2x \right]_{-1}^{0} \]
Evaluating at the limits:
\[ = \left( 0 + 0 + 0 \right) - \left( -\frac{(-1)^3}{3} + \frac{(-1)^2}{2} + 2(-1) \right) \]
\[ = 0 - \left( \frac{1}{3} + \frac{1}{2} - 2 \right) = 0 - \left( \frac{1}{3} + \frac{3}{6} - \frac{12}{6} \right) \]
\[ = 0 - \left( \frac{1}{3} - \frac{9}{6} \right) = 0 - \left( \frac{1}{3} - \frac{3}{2} \right) = 0 - \left( \frac{1}{3} - \frac{9}{6} \right) = 0 - \left( \frac{1}{3} - \frac{3}{2} \right) \]
Calculating gives:
\[ = \frac{1}{6} \]

### Step 5: Calculate the second integral
\[ \int_{0}^{1} (-x + 2 - x^2) \, dx = \int_{0}^{1} (-x^2 - x + 2) \, dx \]
Calculating this integral:
\[ = \left[ -\frac{x^3}{3} - \frac{x^2}{2} + 2x \right]_{0}^{1} \]
Evaluating at the limits:
\[ = \left( -\frac{1}{3} - \frac{1}{2} + 2 \right) - \left( 0 \right) \]
\[ = -\frac{1}{3} - \frac{3}{6} + \frac{12}{6} = -\frac{1}{3} + \frac{9}{6} = -\frac{1}{3} + \frac{3}{2} \]
Calculating gives:
\[ = \frac{1}{6} \]

### Step 6: Total area
The total area is:
\[ A = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3} \]

### Final Result
Thus, the area bounded between the curves \( y = x^2 \) and \( y = -|x| + 2 \) is \( \frac{1}{3} \).