The area enclosed between the curves `y^(2) = x and y= abs(x)` is :

Options

A. `2//3`
B. 1
C. `1//6`
D. `1//3`

Answer

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

### Step 1: Identify the curves
The first curve is \( y^2 = x \), which represents a rightward-opening parabola. The second curve \( y = |x| \) 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 these curves intersect, we need to solve the equations simultaneously.

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

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

### Step 3: Set up the integral for the area
The area \( A \) enclosed between the curves from \( x = 0 \) to \( x = 1 \) can be found using the integral:
\[ A = \int_{0}^{1} (y_{\text{top}} - y_{\text{bottom}}) \, dx \]
In this case, for \( 0 \leq x \leq 1 \):
- The upper curve is \( y = x \)
- The lower curve is \( y = \sqrt{x} \)

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

### Step 4: Calculate the integral
Now we will solve the integral:
\[ A = \int_{0}^{1} (x - \sqrt{x}) \, dx = \int_{0}^{1} x \, dx - \int_{0}^{1} \sqrt{x} \, dx \]

Calculating each integral:
1. \( \int_{0}^{1} x \, dx = \left[ \frac{x^2}{2} \right]_{0}^{1} = \frac{1^2}{2} - 0 = \frac{1}{2} \)
2. \( \int_{0}^{1} \sqrt{x} \, dx = \int_{0}^{1} x^{1/2} \, dx = \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{1} = \frac{2}{3} \)

Putting it all together:
\[ A = \frac{1}{2} - \frac{2}{3} \]

### Step 5: Simplify the result
To combine these fractions:
\[ A = \frac{3}{6} - \frac{4}{6} = \frac{-1}{6} \]
Since area cannot be negative, we take the absolute value:
\[ A = \frac{1}{6} \]

### Final Answer
The area enclosed between the curves \( y^2 = x \) and \( y = |x| \) is \( \frac{1}{6} \).

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