Using integration, find the area of the following region: `R={(x,y):x^2leyle|x|}`

Answer

To find the area of the region \( R = \{(x,y) : x^2 \leq y \leq |x|\} \), we will follow these steps:

### Step 1: Identify the curves
The region is bounded by two curves:
1. \( y = x^2 \) (a parabola opening upwards)
2. \( y = |x| \) (a V-shaped graph)

### Step 2: Find the points of intersection
To find the points where these curves intersect, we set:
\[ x^2 = |x| \]

This gives us two cases to consider:
- Case 1: \( x \geq 0 \) (where \( |x| = x \))
\[ x^2 = x \implies x^2 - x = 0 \implies x(x - 1) = 0 \]
Thus, \( x = 0 \) or \( x = 1 \).

- Case 2: \( x < 0 \) (where \( |x| = -x \))
\[ x^2 = -x \implies x^2 + x = 0 \implies x(x + 1) = 0 \]
Thus, \( x = 0 \) or \( x = -1 \).

The points of intersection are \( (0, 0) \), \( (1, 1) \), and \( (-1, 1) \).

### Step 3: Set up the integral for the area
The area \( A \) can be calculated by integrating the difference between the upper curve and the lower curve from the leftmost intersection point to the rightmost intersection point. The area can be split into two parts due to symmetry.

For \( x \) from \( 0 \) to \( 1 \):
- Upper curve: \( y = |x| = x \)
- Lower curve: \( y = x^2 \)

Thus, the area from \( 0 \) to \( 1 \) is:
\[ A_1 = \int_0^1 (x - x^2) \, dx \]

For \( x \) from \( -1 \) to \( 0 \):
- Upper curve: \( y = |x| = -x \)
- Lower curve: \( y = x^2 \)

Thus, the area from \( -1 \) to \( 0 \) is:
\[ A_2 = \int_{-1}^0 (-x - x^2) \, dx \]

### Step 4: Calculate the area
Calculating \( A_1 \):
\[ A_1 = \int_0^1 (x - x^2) \, dx = \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_0^1 = \left( \frac{1^2}{2} - \frac{1^3}{3} \right) - \left( 0 - 0 \right) = \frac{1}{2} - \frac{1}{3} = \frac{3}{6} - \frac{2}{6} = \frac{1}{6} \]

Calculating \( A_2 \):
\[ A_2 = \int_{-1}^0 (-x - x^2) \, dx = \left[ -\frac{x^2}{2} - \frac{x^3}{3} \right]_{-1}^0 = \left( 0 - 0 \right) - \left( -\frac{(-1)^2}{2} - \frac{(-1)^3}{3} \right) = 0 - \left( -\frac{1}{2} + \frac{1}{3} \right) \]
\[ = \frac{1}{2} - \frac{1}{3} = \frac{3}{6} - \frac{2}{6} = \frac{1}{6} \]

### Step 5: Total area
The total area \( A \) is:
\[ A = A_1 + A_2 = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3} \]

Thus, the area of the region \( R \) is \( \frac{1}{3} \).