Area bounded by the region
`R-={(x,y):y^(2)lexle|y|}` is

Options

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

Explanation

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

### Step 1: Understand the inequalities
The inequalities given are:
1. \( y^2 \leq x \) which can be rewritten as \( x \geq y^2 \).
2. \( x \leq |y| \) which can be rewritten as \( x \leq y \) and \( x \leq -y \).

### Step 2: Graph the equations
We will plot the curves defined by these inequalities:
- The curve \( y^2 = x \) is a parabola opening to the right.
- The lines \( x = y \) and \( x = -y \) are straight lines at 45 degrees to the axes.

### Step 3: Find the intersection points
To find the area, we need to determine the points where these curves intersect. We set \( y^2 = x \) equal to \( x = y \) and \( x = -y \).

1. For \( y^2 = y \):
\[ y^2 - y = 0 \implies y(y - 1) = 0 \implies y = 0 \text{ or } y = 1 \]
Thus, the intersection points are \( (0, 0) \) and \( (1, 1) \).

2. For \( y^2 = -y \):
\[ y^2 + y = 0 \implies y(y + 1) = 0 \implies y = 0 \text{ or } y = -1 \]
Thus, the intersection points are \( (0, 0) \) and \( (1, -1) \).

### Step 4: Set up the integral for the area
The area bounded by the curves can be calculated as:
\[ \text{Area} = 2 \int_0^1 (y - y^2) \, dy \]
This accounts for the area above the x-axis and below the x-axis, which are symmetric.

### Step 5: Calculate the integral
Now we compute the integral:
\[ \int_0^1 (y - y^2) \, dy = \int_0^1 y \, dy - \int_0^1 y^2 \, dy \]
Calculating each part:
1. \( \int_0^1 y \, dy = \left[ \frac{y^2}{2} \right]_0^1 = \frac{1^2}{2} - 0 = \frac{1}{2} \)
2. \( \int_0^1 y^2 \, dy = \left[ \frac{y^3}{3} \right]_0^1 = \frac{1^3}{3} - 0 = \frac{1}{3} \)

Thus, we have:
\[ \int_0^1 (y - y^2) \, dy = \frac{1}{2} - \frac{1}{3} = \frac{3}{6} - \frac{2}{6} = \frac{1}{6} \]

### Step 6: Multiply by 2 for the total area
Since we need to account for both the upper and lower parts:
\[ \text{Total Area} = 2 \times \frac{1}{6} = \frac{1}{3} \]

### Final Answer
The area bounded by the region \( R \) is \( \frac{1}{3} \).