Problem 3 - Calculus Single Variable

Question

In each of Exercises \(1-12,\) calculate the average value of the given function on the given interval. $$ f(x)=1 / x \quad I=[1,4] $$

Step-by-Step Solution

Answer

The average value is \( \frac{1}{3} \ln(4) \).

Step 1: Identify the Formula

The formula for the average value of a function \( f(x) \) on the interval \([a, b]\) is given by \( \text{Average} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx \). In this problem, our function is \( f(x) = \frac{1}{x} \) and the interval is \([1, 4]\).

Step 2: Set Up the Integral

Use the formula identified in Step 1. Substitute \( a = 1 \) and \( b = 4 \) into the formula: \[ \text{Average} = \frac{1}{4-1} \int_{1}^{4} \frac{1}{x} \, dx \]. This simplifies to: \[ \text{Average} = \frac{1}{3} \int_{1}^{4} \frac{1}{x} \, dx \].

Step 3: Evaluate the Indefinite Integral

The indefinite integral of \( \frac{1}{x} \) is \( \ln |x| + C \). However, since we have definite bounds, we don't need the constant \( C \).

Step 4: Evaluate the Definite Integral

Apply the evaluation of the integral from 1 to 4: \[ \int_{1}^{4} \frac{1}{x} \, dx = \left[ \ln |x| \right]_1^4 = \ln(4) - \ln(1) \]. Since \( \ln(1) = 0 \), this simplifies to \( \ln(4) \).

Step 5: Calculate the Average Value

Substitute the result from Step 4 into the expression from Step 2: \[ \text{Average} = \frac{1}{3} (\ln 4) \]. This is the average value of the function on the given interval.

Key Concepts

Definite Integral

When we talk about a definite integral, we are introducing a way to calculate the net area" under a curve. Unlike an indefinite integral, which represents a family of functions and comes with a constant C, a definite integral comes with specific bounds \([a, b]\). These bounds define the interval over which we want to calculate the area.

Here's why definite integrals are useful:

• Calculate the total accumulation of a quantity, like distance traveled.
• Help find the average value of functions over an interval.
• Provide precise calculations by considering exact limits of integration.

In the average value problem, we used the definite integral from 1 to 4 to calculate the specific area under the curve \(f(x) = \frac{1}{x}\), which helped us determine the function's average value over that range.

Continuous Functions

Functions that are continuous are critical in calculus, especially when we are finding integrals. A continuous function is one that has no breaks, jumps, or holes in its graph over its domain. This continuity is essential because the integral of a function is defined as the limiting process of summing the areas of rectangles under the curve. If a function had breaks or discontinuities, calculating the integral would be difficult.

Here are some key points about continuous functions that are useful:

• You can draw them without lifting your pen from the paper.
• They guarantee that approaches involving limits work seamlessly.
• Most elementary functions, like polynomials, exponentials, and logarithms, are continuous over their domains.

In our problem, the function \(f(x) = \frac{1}{x}\) is continuous on the interval \([1, 4]\), qualifying it nicely for integration over this interval.

Integral Evaluation

Evaluating an integral is the process of finding the precise value of a definite integral. This involves finding a function's antiderivative and evaluating it at the upper and lower bounds of the integral. In this particular problem, finding the integral of \(\frac{1}{x}\) required knowing that its antiderivative is \(\ln |x|\).

Steps to evaluate a definite integral include:

• Find the antiderivative of the function you are integrating.
• Substitute the upper and lower boundary values into this antiderivative.
• Calculate the difference between these two results.

During integral evaluation, constants of integration are not needed because the bounds of the definite interval cancel them out. In this exercise, we specifically evaluated \(\ln(x)\) from 1 to 4, concluding that \(\ln(4) - \ln(1) = \ln(4)\). This result then played a crucial role in determining the average value of the function over the interval.