The area bounded by `y=x^(3)-4x` and X-axis is

Options

A. 8
B. 16
C. 4
D. 2

Answer

To find the area bounded by the curve \( y = x^3 - 4x \) and the x-axis, we will follow these steps:

### Step 1: Find the points of intersection with the x-axis
To find where the curve intersects the x-axis, we set \( y = 0 \):

\[ x^3 - 4x = 0 \]

Factoring out \( x \):

\[ x(x^2 - 4) = 0 \]

This gives us:

\[ x(x - 2)(x + 2) = 0 \]

Thus, the roots are \( x = 0, x = 2, x = -2 \). These points are where the curve intersects the x-axis.

### Step 2: Set up the integral for the area
The area under the curve can be calculated by integrating the function from the leftmost intersection point to the rightmost intersection point. Since the curve is above the x-axis between \( -2 \) and \( 0 \) and below the x-axis between \( 0 \) and \( 2 \), we need to calculate the area in two parts:

\[ \text{Area} = \int_{-2}^{0} (x^3 - 4x) \, dx + \int_{0}^{2} -(x^3 - 4x) \, dx \]

### Step 3: Calculate the first integral
Calculating the first integral:

\[ \int (x^3 - 4x) \, dx = \frac{x^4}{4} - 2x^2 \]

Now, we evaluate this from \( -2 \) to \( 0 \):

\[ \left[ \frac{x^4}{4} - 2x^2 \right]_{-2}^{0} = \left( \frac{0^4}{4} - 2(0^2) \right) - \left( \frac{(-2)^4}{4} - 2(-2)^2 \right) \]

Calculating the second part:

\[ = 0 - \left( \frac{16}{4} - 2(4) \right) = 0 - (4 - 8) = 0 + 4 = 4 \]

### Step 4: Calculate the second integral
Now, we calculate the second integral:

\[ \int (-(x^3 - 4x)) \, dx = \int (-x^3 + 4x) \, dx = -\frac{x^4}{4} + 2x^2 \]

Now, we evaluate this from \( 0 \) to \( 2 \):

\[ \left[ -\frac{x^4}{4} + 2x^2 \right]_{0}^{2} = \left( -\frac{2^4}{4} + 2(2^2) \right) - \left( -\frac{0^4}{4} + 2(0^2) \right) \]

Calculating:

\[ = \left( -\frac{16}{4} + 8 \right) - 0 = (-4 + 8) = 4 \]

### Step 5: Add both areas
Now, we add the areas from both integrals:

\[ \text{Total Area} = 4 + 4 = 8 \]

Thus, the area bounded by the curve \( y = x^3 - 4x \) and the x-axis is:

\[ \boxed{8} \text{ square units} \]

---