3.7 Q-9

Question Statement

Find the area between the curve $y = x(x - 1)(x + 1)$ and the $x$-axis.

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Background and Explanation

To solve this problem, we need to calculate the area bounded by the curve and the $x$-axis. This involves finding the points where the curve intersects the $x$-axis, setting up definite integrals, and solving for the area. The curve is a cubic polynomial, so it will change direction, and we need to consider these changes for different intervals.

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Solution

Step 1: Find the points of intersection

We first find the $x$-coordinates where the curve intersects the $x$-axis, i.e., where $y = 0$.

$$y = x(x - 1)(x + 1)$$

Set $y = 0$:

$$x(x - 1)(x + 1) = 0$$

This equation is satisfied when:

$$x = 0, \quad x = 1, \quad x = -1$$

Thus, the curve intersects the $x$-axis at the points $(0, 0)$, $(-1, 0)$, and $(1, 0)$.

Step 2: Determine the regions

Next, we need to identify where the curve is above or below the $x$-axis:

• For $-1 \leq x \leq 0$, $y \geq 0$, so the curve is above the $x$-axis.
• For $0 \leq x \leq 1$, $y \leq 0$, so the curve is below the $x$-axis.

Step 3: Set up the integrals

The total area is the sum of the areas between the curve and the $x$-axis on the two intervals. We calculate the area on each interval separately:

• From $x = -1$ to $x = 0$, the curve is above the $x$-axis.
• From $x = 0$ to $x = 1$, the curve is below the $x$-axis.

The total area is:

$$\text{Area} = \int_{-1}^{0} \left( x^3 - x \right) dx - \int_{0}^{1} \left( x^3 - x \right) dx$$

Step 4: Solve the integrals

For the first integral:

$$\int_{-1}^{0} \left( x^3 - x \right) dx$$

We calculate the integrals of $x^3$ and $-x$ separately:

$$\int x^3 dx = \frac{x^4}{4}, \quad \int x dx = \frac{x^2}{2}$$

Evaluating this from $x = -1$ to $x = 0$:

$$\left[\frac{x^4}{4}\right]_{-1}^{0} - \left[\frac{x^2}{2}\right]_{-1}^{0} = \left(\frac{0^4}{4} - \frac{(-1)^4}{4}\right) - \left(\frac{0^2}{2} - \frac{(-1)^2}{2}\right)$$ $$= \left(0 - \frac{1}{4}\right) - \left(0 - \frac{1}{2}\right) = -\frac{1}{4} + \frac{1}{2} = \frac{1}{4}$$

For the second integral:

$$\int_{0}^{1} \left( x^3 - x \right) dx$$

Similarly, we calculate the integrals of $x^3$ and $-x$ separately:

$$\int x^3 dx = \frac{x^4}{4}, \quad \int x dx = \frac{x^2}{2}$$

Evaluating this from $x = 0$ to $x = 1$:

$$\left[\frac{x^4}{4}\right]_{0}^{1} - \left[\frac{x^2}{2}\right]_{0}^{1} = \left(\frac{1^4}{4} - \frac{0^4}{4}\right) - \left(\frac{1^2}{2} - \frac{0^2}{2}\right)$$ $$= \left(\frac{1}{4} - 0\right) - \left(\frac{1}{2} - 0\right) = \frac{1}{4} - \frac{1}{2} = -\frac{1}{4}$$

Step 5: Combine the results

Now, we subtract the two areas:

$$\text{Total Area} = \frac{1}{4} - \left(-\frac{1}{4}\right) = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$$

Thus, the area between the curve and the $x$-axis is $\frac{1}{2}$ square units.

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Key Formulas or Methods Used

• Definite integrals: To calculate the area between a curve and the $x$-axis.

$$\text{Area} = \int_{a}^{b} f(x) , dx$$

• Factorization: Used to find the points where the curve intersects the $x$-axis by solving $f(x) = 0$.

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Summary of Steps

1. Find the points of intersection: Solve $x(x - 1)(x + 1) = 0$ to get $x = -1, 0, 1$.
2. Determine the regions: Identify where the curve is above or below the $x$-axis.
3. Set up the integrals: Divide the area into two integrals, one for $-1 \leq x \leq 0$ and one for $0 \leq x \leq 1$.
4. Solve the integrals: Compute the integrals and find the area for each region.
5. Combine the results: Add the areas to get the total area, which is $\frac{1}{2}$ square units.