Problem 30 - Applied Calculus

Question

Find the exact area of the region bounded by the \(x\) -axis and the graph of \(y=x^{3}-x\).

Step-by-Step Solution

Answer

The exact area is \( \frac{1}{2} \).

Step 1: Determine the Intersection Points

First, find the points where the graph of the function intersects the x-axis. Set the equation \( y=x^3-x \) to zero: \[ x^3 - x = 0 \]Factoring gives:\[ x(x^2 - 1) = 0 \]Further factor to find the roots:\[ x(x-1)(x+1) = 0 \]So the intersection points are \( x = -1, 0, \) and \( 1 \).

Step 2: Set Up the Integral

Since we need the area between \( y = x^3 - x \) and the x-axis from \( x = -1 \) to \( x = 1 \), we integrate the absolute value of the function over the interval:\[ \int_{-1}^{0} -(x^3 - x)\,dx + \int_{0}^{1} (x^3 - x)\,dx \]

Step 3: Evaluate the Integral from -1 to 0

Integrate \( -(x^3 - x) \) from \( x = -1 \) to \( 0 \):\[ \int_{-1}^{0} (-x^3 + x) \, dx = \left[ \frac{-x^4}{4} + \frac{x^2}{2} \right]_{-1}^{0} \]Evaluating gives:\[ \left( 0 - 0 \right) - \left( \frac{-(-1)^4}{4} + \frac{(-1)^2}{2} \right) = 0 - \left( \frac{-1}{4} + \frac{1}{2} \right) = \frac{-1}{4} + \frac{2}{4} = \frac{1}{4}\]

Step 4: Evaluate the Integral from 0 to 1

Integrate \( x^3 - x \) from \( x = 0 \) to \( 1 \):\[\int_{0}^{1} (x^3 - x) \, dx = \left[ \frac{x^4}{4} - \frac{x^2}{2} \right]_{0}^{1} \]Evaluating gives:\[\left( \frac{1^4}{4} - \frac{1^2}{2} \right) - \left( 0 - 0 \right) = \frac{1}{4} - \frac{1}{2} = \frac{1}{4} - \frac{2}{4} = \frac{-1}{4}\]

Step 5: Add the Absolute Values of the Areas Found

Add the absolute values of the integrals from steps 3 and 4 to find the total area:\[\frac{1}{4} + \left| \frac{-1}{4} \right| = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2}\]

Step 6: Conclusion

The exact area of the region bounded by the x-axis and the graph of \( y = x^3 - x \) from \( x = -1 \) to \( x = 1 \) is \( \frac{1}{2} \).

Key Concepts

Area Between Curves

When solving problems involving the area between curves, it's important to understand which parts of the graph are above and below the x-axis. This can help in setting up the correct integral to find the desired area. The concept revolves around calculating areas using definite integrals.
For the function given, the area between the curve and the x-axis is determined by calculating separate integral sections where the function is both above and below the x-axis.
If a function is entirely above the x-axis, simply integrate the function over the interval. If the function dips below the axis, take the absolute value of the function where necessary to ensure you calculate the area correctly.

• Identify intervals where the function crosses the x-axis.
• Use the definite integral to calculate the area of each section.
• Employ absolute values if parts of the curve lie below the x-axis.

By integrating over these intervals and adding the absolute values, you determine the total area between the curve and the x-axis.

Intersection Points

Finding intersection points is step one in solving area problems, as these points determine the limits of integration.
In the given example, the function intersects the x-axis where its value is zero. This results in setting the function equal to zero and solving for x.
For the function \( y = x^3 - x \), solve:

• Set the equation to zero: \(x^3 - x = 0\).
• Factor the polynomial to find where it equals zero.
• Find that \(x = -1, 0,\) and \(1\) are the intersection points.

These points define the sections of the graph where the area under the curve needs to be calculated, thus serving as the bounds for your integrals.

Factoring Polynomials

Factoring polynomials is essential for finding the roots of a function, which can be used to identify important features like intersection points.
In this case, the polynomial \(x^3 - x\) was set to zero and solved by factoring.
Here's how the process works:

• Factor out the greatest common factor, here it's \(x\).
• After factoring, you're left with \(x(x^2 - 1)\).
• Further factor \(x^2 - 1\) into \((x-1)(x+1)\).

This gives the solutions \(x = -1, 0,\) and \(1\). Factoring helps simplify the expression and reveals key points for setting up integrals.

Absolute Value

The concept of absolute value is crucial when dealing with definite integrals of functions below the x-axis, as it ensures the area calculation remains positive.
In our exercise, between the interval \([-1, 0]\), the function is negative, so we take the absolute value of the integral over this interval.

• Recognize when and where the function is below the x-axis by checking the sign of the function value in the interval.
• When integrating, use the absolute value of the function to compute the area accurately.
• Apply this understanding to ensure the entire area is accounted for positively before summing the segments.

Adopting the absolute value in integration ensures that each part of the curve contributes positively to the total area calculation.