Problem 58 - Applied Calculus

Question

Find the area bounded by the given curves. \(y=x^{n}\) and \(y=x^{n-1}(\) for \(n>1)\)

Step-by-Step Solution

Answer

The area is \(\frac{1}{n(n+1)}\).

Step 1: Determine Intersection Points

The area is bounded by the curves where they intersect. Set the two equations equal to find these points: \[ x^n = x^{n-1} \]Divide both sides by \(x^{n-1}\) (assuming \(x eq 0\)):\[ x = 1 \]Thus, the curves intersect at \(x = 1\). Additionally, since \(x = 0\) satisfies both equations, the other point of intersection is \(x = 0\).

Step 2: Set Up the Integral for Area

To find the area between the two curves from \(x = 0\) to \(x = 1\), compute the integral of the difference between the two functions:\[ A = \int_{0}^{1} (x^{n-1} - x^n) \, dx \]

Step 3: Find the Integral

Calculate the integral:\[ A = \int_{0}^{1} (x^{n-1} - x^n) \, dx = \int_{0}^{1} x^{n-1} \, dx - \int_{0}^{1} x^n \, dx \]Let's find each integral separately:\[ \int x^{n-1} \, dx = \frac{x^n}{n} \]and\[ \int x^n \, dx = \frac{x^{n+1}}{n+1} \]

Step 4: Evaluate the Definite Integral

Evaluate the definite integrals from \(x = 0\) to \(x = 1\):\[ \int_{0}^{1} x^{n-1} \, dx = \left[ \frac{x^n}{n} \right]_0^1 = \frac{1^n}{n} - \frac{0^n}{n} = \frac{1}{n} \]\[ \int_{0}^{1} x^n \, dx = \left[ \frac{x^{n+1}}{n+1} \right]_0^1 = \frac{1^{n+1}}{n+1} - \frac{0^{n+1}}{n+1} = \frac{1}{n+1} \]

Step 5: Calculate the Area

Subtract the integrals to find the total area between the curves:\[ A = \frac{1}{n} - \frac{1}{n+1} \]Simplify the expression:\[ A = \frac{(n+1) - n}{n(n+1)} = \frac{1}{n(n+1)} \]

Key Concepts

Definite Integral

The definite integral is a fundamental concept in calculus. It helps us calculate the exact area under a curve over a given interval. In this particular problem, we use it to find the area between two curves, which is pivotal in understanding how these functions interact within a specified range.

To apply a definite integral, we need:

• The functions that define the curves.
• The interval over which we will integrate.

In this problem, the functions are given by the curves \(y=x^n\) and \(y=x^{n-1}\), and the interval is determined by the intersection points \([0, 1]\). The definite integral is expressed as:\[A = \int_{0}^{1} (x^{n-1} - x^n) \, dx\]
This integral sums up the vertical distances between the curves from the point where \(x = 0\) to \(x = 1\), giving us the area between these two elegant curves.

Intersection Points

Finding intersection points is a crucial step when solving problems involving areas between curves. This is because these points define the limits of our integration.

To determine where the curves \(y=x^n\) and \(y=x^{n-1}\) intersect, we set the two equations equal:\[x^n = x^{n-1}\]
Upon solving and simplifying, we find that the curves intersect at \(x = 0\) and \(x = 1\). These values are inside the domain where both functions are defined. Here, \(x = 1\) is found by dividing both sides of the equation by \(x^{n-1}\), considering \(x eq 0\).

These intersection points are critical. They tell us where to start and end our integral, ensuring that we're calculating the area over the correct interval.

Calculus Problem Solving

Solving problems using calculus requires a systematic approach. For finding areas between curves, this involves several key steps.

Here's an easy-to-follow process:

• Identify the functions you are working with.
• Find their intersection points to set the limits of integration.
• Set up the definite integral using the difference between the functions over the intersection interval.
• Calculate the definite integral by solving each part separately if needed.
• Simplify the result to get the total area.

For our problem, after finding intersection points at \(x = 0\) and \(x = 1\), we set up the integral:\[\int_{0}^{1} (x^{n-1} - x^n) \, dx\]
Then, we evaluate the integrals separately:\[\int_{0}^{1} x^{n-1} \, dx \quad \text{and} \quad \int_{0}^{1} x^n \, dx\]
Finally, we subtract these integrals to find the total area. By following these steps carefully, we reach a simplified expression for the area:\[A = \frac{1}{n(n+1)}\]
This structured approach makes tackling calculus problems like these more manageable and less daunting.