Chapter 7: Problem 3
In each of Exercises \(1-12,\) calculate the average value of the given function on the given interval. $$ f(x)=1 / x \quad I=[1,4] $$
Short Answer
Expert verified
The average value is \( \frac{1}{3} \ln(4) \).
Step by step solution
01
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 \([1, 4]\).
02
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: \[ \text{Average} = \frac{1}{3} \int_{1}^{4} \frac{1}{x} \, dx \].
03
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 \( C \).
04
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 \( \ln(4) \).
05
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.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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:
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.
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:
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.
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:
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.