The area of the region bounded by the curve `y = x^(2)` and y `= |x|` is equal to

Options

A. `(5)/(3)` sq. units
B. `(1)/(3)` sq. units
C. `(5)/(3)` sq. units
D. `(1)/(6)` sq. unit

Answer

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

### Step 1: Identify the curves
The first curve is \( y = x^2 \), which is a parabola opening upwards. The second curve is \( y = |x| \), which consists of two lines: \( y = x \) for \( x \geq 0 \) and \( y = -x \) for \( x < 0 \).

### Step 2: Find the points of intersection
To find the points where the curves intersect, we need to solve the equations \( x^2 = x \) and \( x^2 = -x \).

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

2. For \( y = -x \):
\[ x^2 = -x \implies x^2 + x = 0 \implies x(x + 1) = 0 \]
This gives us \( x = 0 \) 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 difference between the upper curve and the lower curve.

- For \( x \) in the interval \([-1, 0]\), the upper curve is \( y = -x \) and the lower curve is \( y = x^2 \).
- For \( x \) in the interval \([0, 1]\), the upper curve is \( y = x \) and the lower curve is \( y = x^2 \).

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

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

### Step 5: Combine the areas
The total area \( A \) is:
\[ A = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3} \]

### Final Answer
The area of the region bounded by the curves \( y = x^2 \) and \( y = |x| \) is \( \frac{1}{3} \) square units.

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