Problem 42 - Calculus and its applications

Question

Find the average function value over the given interval. $$ f(x)=x^{n}, n \neq 0 ; \quad[0,1] $$

Step-by-Step Solution

Answer

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

Step 1: Understanding the Formula for Average Function Value

The average value of a function \( f(x) \) over the interval \([a, b]\) is calculated using the formula: \[ \frac{1}{b-a} \int_{a}^{b} f(x) \, dx \]. In this problem, the interval is \([0,1]\), so \( a = 0 \) and \( b = 1 \).

Step 2: Setting Up the Integral

Substitute \( f(x) = x^n \), \( a = 0 \), and \( b = 1 \) into the average value formula. This gives us: \[ \frac{1}{1-0} \int_{0}^{1} x^n \, dx = \int_{0}^{1} x^n \, dx \].

Step 3: Calculating the Integral

To find \( \int_{0}^{1} x^n \, dx \), we use the power rule for integration. For \( x^n \), the integral is \( \frac{x^{n+1}}{n+1} \). Evaluating this from 0 to 1 gives: \[ \left. \frac{x^{n+1}}{n+1} \right|_{0}^{1} = \frac{1^{n+1}}{n+1} - \frac{0^{n+1}}{n+1} = \frac{1}{n+1} \].

Step 4: Conclusion

The average function value for \( f(x) = x^n \) over the interval \([0, 1]\) is \( \frac{1}{n+1} \). This solution highlights the importance of correctly setting up and computing definite integrals.

Key Concepts

Definite Integral

The concept of a definite integral is crucial in calculus. It's about finding the total accumulation of quantities, which can often be represented as the area under a curve of a function between two points, known as the bounds. Definite integrals have fixed starting and ending points, unlike indefinite integrals, which have no specific limits or bounds.
In our exercise, we calculate the definite integral from 0 to 1 of the function \( f(x) = x^n \). This process involves using a particular rule of integration to find the value of this expression between these two points. Remember, when calculating a definite integral, you don't need to add the constant \(+C\) that appears in indefinite integrals. The fundamental theorem of calculus links derivatives and integrals, making it possible to evaluate such expressions without knowing the antiderivative's constant.

Interval Notation

Interval notation is a mathematical shorthand used to describe a range of values. These ranges are typically depicted as pairs of values within brackets. There are different types of brackets used depending on whether the endpoints of the interval are included or excluded.

• Square brackets \([a, b]\) mean the endpoints \(a\) and \(b\) are included in the interval.
• Parentheses \((a, b)\) indicate that the endpoints are not included.

For example, in our exercise, the interval \([0, 1]\) includes both 0 and 1. This means that while calculating the average value of the function, the integral takes into account every point from 0 to 1, including 0 and 1 as well. Understanding this notation allows one to accurately define the domain over which they are performing calculations.

Power Rule for Integration

The power rule for integration is a fundamental technique used to calculate integrals, specifically those involving power functions. The formula is straightforward: for \( x^n \), the integral is calculated as \( \frac{x^{n+1}}{n+1} + C \), given \( n eq -1 \). This formula allows you to find the antiderivative easily.
In the context of the definite integral from the exercise, this rule helps us find \( \int_{0}^{1} x^n \, dx \). We follow through by evaluating \( \left. \frac{x^{n+1}}{n+1} \right|_{0}^{1} \), thereby omitting the constant \(+C\). For a function like \( x^n \), which is a simple polynomial, the power rule simplifies the integral significantly, making it a critical tool in calculus.