3.7 Q-8

Question Statement

Find the area bounded by the curve $y = x^3 - 4x$ and the $x$-axis.

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Background and Explanation

In this problem, we need to find the area between the curve $y = x^3 - 4x$ and the $x$-axis. To solve this, we will:

1. Identify the points where the curve intersects the $x$-axis (i.e., where $y = 0$).
2. Set up integrals for the areas above and below the $x$-axis and compute them separately.

The curve intersects the $x$-axis where the equation $y = x^3 - 4x$ equals zero. We will use this information to divide the problem into parts and find the area.

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Solution

Step 1: Find the x-intercepts

The curve intersects the $x$-axis when $y = 0$, so set $x^3 - 4x = 0$:

$$x(x^2 - 4) = 0$$

This gives two factors:

• $x = 0$
• $x^2 - 4 = 0$, which simplifies to $x = \pm 2$.

So, the curve cuts the $x$-axis at the points $(-2, 0)$, $(0, 0)$, and $(2, 0)$.

Step 2: Determine the sign of $y$ in the intervals

• For $-2 \leq x \leq 0$, we have $y \geq 0$, so the curve is above the $x$-axis.
• For $0 \leq x \leq 2$, we have $y \leq 0$, so the curve is below the $x$-axis.

Step 3: Set up the integral

The area can be split into two parts:

• From $x = -2$ to $x = 0$, the curve is above the $x$-axis, so the area is positive.
• From $x = 0$ to $x = 2$, the curve is below the $x$-axis, so the area is negative (we will subtract the integral to account for the negative region).

Thus, the total area is:

$$A = \int_{-2}^{0} y , dx - \int_{0}^{2} y , dx$$

Substitute $y = x^3 - 4x$ into both integrals:

$$A = \int_{-2}^{0} (x^3 - 4x) , dx - \int_{0}^{2} (x^3 - 4x) , dx$$

Step 4: Compute the integrals

We split the integrals into individual terms:

$$A = \left( \int_{-2}^{0} x^3 , dx - 4 \int_{-2}^{0} x , dx \right) - \left( \int_{0}^{2} x^3 , dx - 4 \int_{0}^{2} x , dx \right)$$

Now, compute each integral:

• The integral of $x^3$ is $\frac{x^4}{4}$.
• The integral of $x$ is $\frac{x^2}{2}$.

Thus, we evaluate each part:

1. From $x = -2$ to $x = 0$:

$$\int_{-2}^{0} x^3 , dx = \left[ \frac{x^4}{4} \right]_{-2}^{0} = \frac{(0)^4 - (-2)^4}{4} = -\frac{16}{4} = -4$$ $$\int_{-2}^{0} x , dx = \left[ \frac{x^2}{2} \right]_{-2}^{0} = \frac{(0)^2 - (-2)^2}{2} = -\frac{4}{2} = -2$$

Thus, for the first part of the area:

$$\left( -4 - 4(-2) \right) = -4 + 8 = 4$$

2. From $x = 0$ to $x = 2$:

$$\int_{0}^{2} x^3 , dx = \left[ \frac{x^4}{4} \right]_{0}^{2} = \frac{(2)^4 - (0)^4}{4} = \frac{16}{4} = 4$$ $$\int_{0}^{2} x , dx = \left[ \frac{x^2}{2} \right]_{0}^{2} = \frac{(2)^2 - (0)^2}{2} = \frac{4}{2} = 2$$

Thus, for the second part of the area:

$$\left( 4 - 4(2) \right) = 4 - 8 = -4$$

Step 5: Final calculation

Now, subtract the two results:

$$A = 4 - (-4) = 4 + 4 = 8$$

Thus, the total area is:

$$A = 8 , \text{square units}$$

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Key Formulas or Methods Used

• Definite Integral: The area under the curve between $x = a$ and $x = b$ is given by:

$$A = \int_a^b f(x) , dx$$

• Power Rule of Integration:

$$\int x^n , dx = \frac{x^{n+1}}{n+1}$$

• Splitting the Integral: The area is divided into regions where the function is above and below the $x$-axis, and we handle each part separately.

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Summary of Steps

1. Find the $x$-intercepts by solving $x^3 - 4x = 0$, which gives $x = -2, 0, 2$.
2. Determine the sign of $y$ in the intervals $[-2, 0]$ and $[0, 2]$.
3. Set up the integrals for the area:

$$A = \int_{-2}^{0} (x^3 - 4x) , dx - \int_{0}^{2} (x^3 - 4x) , dx$$

4. Compute the integrals for both parts: $A = 4 - (-4) = 8$
5. The total area is: $A = 8 , \text{square units}$