Problem 21 - Applied Calculus

Question

Find the average value of each function over the given interval. \(f(x)=x^{n}\) on \([0,1],\) where \(n\) is a constant \((n>0)\)

Step-by-Step Solution

Answer

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

Step 1: Understand the Average Value Formula

The average value of a continuous function \(f(x)\) over the interval \([a, b]\) is given by \[\frac{1}{b-a} \int_{a}^{b} f(x) \, dx.\]Here, \(a = 0\) and \(b = 1\), so the formula simplifies to \[\int_{0}^{1} f(x) \, dx.\]

Step 2: Set Up the Integral

We need to find the average value of \(f(x) = x^n\) over \([0, 1]\). Set up the integral:\[\frac{1}{1-0} \int_{0}^{1} x^n \, dx = \int_{0}^{1} x^n \, dx.\]

Step 3: Integrate the Function

To evaluate the integral \(\int_{0}^{1} x^n \, dx\), use the power rule for integration, which states that \[\int x^n \, dx = \frac{x^{n+1}}{n+1} + C.\]Since we are evaluating a definite integral, we don't need the constant \(C\). Apply the limits:\[\left. \frac{x^{n+1}}{n+1} \right|_{0}^{1}.\]

Step 4: Evaluate the Integral at the Limits

Substitute the limits into the evaluated integral:\[\frac{1^{n+1}}{n+1} - \frac{0^{n+1}}{n+1} = \frac{1}{n+1} - 0.\]

Step 5: Conclusion

The average value of the function \(f(x) = x^n\) over the interval \([0, 1]\) is \(\frac{1}{n+1}.\)

Key Concepts

Continuous Function

A continuous function is a fundamental concept in calculus and analysis. Imagine drawing a curve of the function on a graph without lifting your pen. This smoothness means there are no gaps, jumps, or sharp turns.
A continuous function's value changes very predictably and smoothly over its domain. This predictability is important because it allows us to perform operations like integration and differentiation with confidence.

For a function to be continuous at a point:

• The function must be defined at that point.
• The limit as it approaches the point from both directions must exist.
• The function's value at the point must equal the limit of the function as it approaches the point.

In our exercise, the function \(f(x) = x^n\) is continuous over any real number interval, including \[0, 1\]. This assures us that we can calculate the average value using integration over this range without encountering mathematical interruptions.

Definite Integral

The definite integral is like the king of integration. It gives us the accumulated total of a function over a specific interval, providing a sort of 'sum' of what the function describes, considering both area and direction.
In mathematical terms, the definite integral is written as \[ \int_{a}^{b} f(x) \, dx \]. This notation means: calculate the integral of \(f(x)\) from \(x = a\) to \(x = b\).

During this exercise, integrating \(f(x) = x^n\) from \([0,1]\) computes the area under the curve of \(x^n\) between those two points on the x-axis, capturing the essence of the function over that range. The result helps us find the average value of the continuous function across the interval, which is also defined as \[ \frac{1}{b-a} \int_{a}^{b} f(x) \, dx \]. This formula confirms that the contribution of each part of the interval to the total is equally considered, balancing out peaks and low points.

Power Rule for Integration

The power rule for integration is a handy shortcut for integrating polynomial functions. It simplifies the calculation of integrals, especially when dealing with powers of x.
The rule states: if you want to integrate \(x^n\) where eq -1\, the result is \[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \]. Here, \C\ is the constant of integration for indefinite integrals, representing the 'family' of antiderivatives.

For definite integrals, the constant \C\ isn't needed. After finding the antiderivative, we substitute the upper and lower limits of the interval directly into this expression. For our task, when solving \(\int_{0}^{1} x^n \, dx\), after applying the power rule, we replace \(x\) with 1 and 0 to get \[ \left. \frac{x^{n+1}}{n+1} \right|_{0}^{1} = \frac{1^{n+1}}{n+1} - \frac{0^{n+1}}{n+1} = \frac{1}{n+1} - 0 = \frac{1}{n+1} \].This process turns a potentially complex problem into a simple plug-and-play solution, demonstrating the beauty and utility of calculus in solving real-world and theoretical problems.