The area bounded by the curve `y=x(x^(2)-1)` and x-axis is

Options

A. 0
B. `(1)/(2)`
C. 1
D. `(1)/(4)`

Answer

To find the area bounded by the curve \( y = x(x^2 - 1) \) and the x-axis, we can follow these steps:

### Step 1: Identify the curve and find the points of intersection with the x-axis.
The curve is given by:
\[ y = x(x^2 - 1) = x^3 - x \]
To find the points where the curve intersects the x-axis, we set \( y = 0 \):
\[ x(x^2 - 1) = 0 \]
This gives us:
\[ x = 0 \quad \text{or} \quad x^2 - 1 = 0 \implies x = 1 \quad \text{or} \quad x = -1 \]
Thus, the points of intersection are \( x = -1, 0, 1 \).

### Step 2: Determine the area under the curve.
The area bounded by the curve and the x-axis from \( x = -1 \) to \( x = 1 \) can be calculated using integration. Since the curve is symmetric about the y-axis, we can calculate the area from \( x = 0 \) to \( x = 1 \) and then double it.

### Step 3: Set up the integral for the area.
The area \( A \) can be expressed as:
\[ A = 2 \int_0^1 (x^3 - x) \, dx \]

### Step 4: Calculate the integral.
Now we compute the integral:
\[ \int (x^3 - x) \, dx = \int x^3 \, dx - \int x \, dx = \frac{x^4}{4} - \frac{x^2}{2} \]
Evaluating from \( 0 \) to \( 1 \):
\[ \left[ \frac{x^4}{4} - \frac{x^2}{2} \right]_0^1 = \left( \frac{1^4}{4} - \frac{1^2}{2} \right) - \left( \frac{0^4}{4} - \frac{0^2}{2} \right) \]
\[ = \left( \frac{1}{4} - \frac{1}{2} \right) - 0 = \frac{1}{4} - \frac{2}{4} = -\frac{1}{4} \]
Since we are interested in the area, we take the absolute value:
\[ \text{Area from } 0 \text{ to } 1 = \frac{1}{4} \]

### Step 5: Multiply by 2 for the total area.
Thus, the total area is:
\[ A = 2 \times \frac{1}{4} = \frac{1}{2} \]

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