Find the area bounded by curves `{(x,y): y ge x ^(2) and y = |x|}.`

Answer

To find the area bounded by the curves \( y = x^2 \) and \( y = |x| \), we can follow these steps:

### Step 1: Understand the curves
The curve \( y = |x| \) consists of two lines:
- \( y = x \) for \( x \geq 0 \)
- \( y = -x \) for \( x < 0 \)

The curve \( y = x^2 \) is a parabola that opens upwards.

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

1. For \( y = x \):
\[ x^2 = x \implies x^2 - x = 0 \implies x(x - 1) = 0 \]
This gives us the points \( x = 0 \) and \( x = 1 \).

2. For \( y = -x \):
\[ x^2 = -x \implies x^2 + x = 0 \implies x(x + 1) = 0 \]
This gives us the points \( x = 0 \) and \( x = -1 \).

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

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

For \( x \in [-1, 0] \):
- Upper curve: \( y = -x \)
- Lower curve: \( y = x^2 \)

For \( x \in [0, 1] \):
- Upper curve: \( y = x \)
- Lower curve: \( y = x^2 \)

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

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

2. Calculate the second integral:
\[ \int_{0}^{1} (x - x^2) \, dx = \left[\frac{x^2}{2} - \frac{x^3}{3}\right]_{0}^{1} \]
Evaluating this:
\[ = \left[\frac{1^2}{2} - \frac{1^3}{3}\right] - \left[0\right] = \frac{1}{2} - \frac{1}{3} = \frac{3}{6} - \frac{2}{6} = \frac{1}{6} \]

### Step 5: Combine the areas
Now, we combine the areas from both integrals:
\[ A = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3} \]

### Final Answer
The area bounded by the curves \( y = x^2 \) and \( y = |x| \) is:
\[ \boxed{\frac{1}{3}} \text{ square units} \]