Problem 43 - Calculus and its applications

Question

Find the average function value over the given interval. $$ f(x)=x^{n}, n \neq 0 ; \quad[1,2] $$

Step-by-Step Solution

Answer

The average function value of \( f(x) = x^n \) over \([1,2]\) is \( \frac{2^{n+1} - 1}{n+1} \).

Step 1: Understand the Average Function Value Formula

The average value of a function \( f(x) \) over an interval \([a, b]\) is given by the formula \[\frac{1}{b-a} \int_{a}^{b} f(x) \, dx\] This allows us to compute the mean value of \( f(x) \) on the interval \([a, b]\). Here, \( a = 1 \) and \( b = 2 \).

Step 2: Set Up the Integral

For the given function \( f(x) = x^n \), we need to evaluate\[\int_{1}^{2} x^n \, dx\]Substituting into the average formula, we have:\[\frac{1}{2-1} \int_{1}^{2} x^n \, dx = \int_{1}^{2} x^n \, dx\] We will solve this integral to find the average value.

Step 3: Solve the Integral

The antiderivative of \( x^n \), where \( n eq -1 \), is \( \frac{x^{n+1}}{n+1} \). Apply this formula:\[\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\]Evaluate from 1 to 2:\[\left[\frac{x^{n+1}}{n+1}\right]_{1}^{2} = \frac{2^{n+1}}{n+1} - \frac{1^{n+1}}{n+1}\]

Step 4: Simplify the Expression

Simplify the expression obtained:\[\int_{1}^{2} x^n \, dx = \frac{2^{n+1} - 1}{n+1}\]The average function value of \( f(x) \) over \([1,2]\) is:\[\frac{1}{2-1} \times \frac{2^{n+1} - 1}{n+1} = \frac{2^{n+1} - 1}{n+1}\]Therefore, the average value of the function from 1 to 2 is \( \frac{2^{n+1} - 1}{n+1} \).

Key Concepts

Integral Calculus

Integral calculus is one of the two major branches of calculus, alongside differential calculus. It focuses on finding the integral of a function, which essentially helps in determining the total accumulation of quantities like area, volume, or mass. Think of it as adding up all the bits and pieces that make up a shape or a quantity.

Key concepts in integral calculus include:

• Riemann Sums: The method used for approximating the total area under a curve by dividing it into shapes (usually rectangles).
• Definite Integral: This is the main concept which gives the total accumulation over an interval. It's represented as: \[ \int_{a}^{b} f(x) \, dx \]
• Fundamental Theorem of Calculus: This connects the concept of the derivative of a function to the integral, establishing a relationship between differentiation and integration.

Wrapping up, the goal of integral calculus is to determine the size or volume of an object or space when expressed as a function. It's extremely helpful in real-world applications, from physics to engineering, where such calculations are often necessary.

Antiderivative

An antiderivative, also known as an indefinite integral, is a function that reverses differentiation. In other words, it takes you from the derivative back to the original function. When you find an antiderivative, you're finding a family of functions, each differing by a constant, that could have led to the given derivative.

Consider the process like this:

• The operation of finding an antiderivative of a function \( f(x) \) is working backward to determine a function that, when differentiated, gives back \( f(x) \).
• The notation for an antiderivative is the integral sign without limits: \( \int f(x) \, dx \).
• Every antiderivative follows the form \( F(x) + C \), where \( C \) is an arbitrary constant.

For example, if you know \( f(x) = x^n \), the antiderivative is found using the power rule:\[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C, \quad \text{for} \; n eq -1. \]Understanding antiderivatives is crucial in finding definite integrals, which we will explore next.

Definite Integral

A definite integral calculates the accumulation of a quantity, like area under a curve, within a specified interval. Using the definite integral, we can find the exact "size" of this accumulation.

For the given function \( f(x) = x^n \), its definite integral from \( 1 \) to \( 2 \) is represented as:\[ \int_{1}^{2} x^n \, dx. \]
Steps to solve a definite integral include:

• Find the antiderivative, \( \int x^n \, dx = \frac{x^{n+1}}{n+1} \).
• Substitute the bounds of the interval (in this case, 1 and 2) into the antiderivative to find the net accumulation.
• Calculate \( F(b) - F(a) \), where \( F(x) \) is the antiderivative, \( a \) is the lower limit and \( b \) is the upper limit.

For example, in this exercise, the definite integral results in:\[ \left[ \frac{x^{n+1}}{n+1} \right]_{1}^{2} = \frac{2^{n+1}}{n+1} - \frac{1^{n+1}}{n+1}. \]This gives the area or total accumulation for the interval, helping us find the average value of our function over that interval.