2.8: The Mean Value Theorem for Integrals

Section Preview

Note to the Instructor (click to expand)

This is an incredibly short section—perhaps a half lecture.

Core Prerequisites (i.e., "Understanding the Lesson"): The following prerequisite topics (which have not already been listed as prerequisites in previous sections of this text) are required to understand the core concepts in this lesson.

(click to expand)

• Sets and Numbers
  • Inequalities and Inequality Notation: The proof of the theorem hinges on reading and manipulating the compound inequality \(m \leq f(x) \leq M\) (and the derived \(m \leq \frac{1}{b-a}\int_a^b f\,dx \leq M\)); understanding that the average value is "a number between \(m\) and \(M\)" is what licenses the Intermediate Value Theorem step.

Operational Prerequisites (i.e., "Completing the Homework"): The following prerequisite topics (which have not already been listed as prerequisites in previous sections of this text) are needed to complete the homework.

(click to expand)

• Solving Equations
  • Solving Quadratic Equations Using Extraction of Roots: In Example 2, solving \(c^2=3\) requires extracting roots to get \(c=\pm\sqrt{3}\) (then selecting the value inside the interval); Checkpoint 2 uses the identical maneuver.
• Trigonometry
  • Solving Trigonometric Equations Using Identities: Homework problem #9 requires solving a trigonometric equation to find the times when the number of daylight hours equals the average value.

The following is a list of learning objectives for this section.

Learning Objectives (click to expand)

• Describe the meaning of the Mean Value Theorem for Integrals.

This short section is dedicated to an examination of another essential theorem, the Mean Value Theorem for Integrals.

The Mean Value Theorem for Integrals

Recall, for a discrete list of values, \(\{a_1, a_2, \ldots, a_n \}\), we compute the average using the formula\[a_{\text{avg}} = \dfrac{a_1 + a_2 + \cdots + a_n}{n}.\nonumber\]We now have the "mathematical technology" to create the analog for a continuous function, \(f(x)\), over a closed interval \([a,b]\).

We can approximate the average value of \(f\) as follows\[f_{\text{avg}} \approx \dfrac{\sum_{i = 1}^n f(x_i)}{n},\nonumber\]where \(x_i\) is defined using our traditional definitions from Riemann sums. However, we now perform a little manipulation to force the numerator to become a Riemann sum.\[f_{\text{avg}} \approx \dfrac{\Delta x \sum_{i = 1}^n f(x_i)}{n \Delta x} = \dfrac{\sum_{i = 1}^n f(x_i) \Delta x}{n \Delta x} = \dfrac{\sum_{i = 1}^n f(x_i) \Delta x}{b - a},\nonumber\]where we used the fact that \(\Delta x = \frac{b - a}{n}\) in our last step.

Taking the limit as \(n \to \infty\), we obtain the exact value of the average for \(f\) over the closed interval \([a,b]\), simply called the average value of \(f\),\[f_{\text{avg}} = \dfrac{1}{b-a} \int_a^b f(x)\,dx.\nonumber\]

This definition leads us to a new theorem to add to our "toolkit." The Mean Value Theorem for Integrals states that a continuous function on a closed interval takes on its average value at the same point in that interval. The theorem guarantees that if \(f(x)\) is continuous, a point \(c\) exists in an interval \([a,b]\) such that the value of the function at \(c\) is equal to the average value of \(f(x)\) over \([a,b]\). We state this theorem mathematically with the help of the formula for the average value of a function that we presented in Calculus I.

Theorem: The Mean Value Theorem for Integrals

If \(f(x)\) is continuous over an interval \([a,b]\), then there is at least one point \(c \in [a,b]\) such that\[f(c)=\dfrac{1}{b-a} \int ^b_af(x)\,dx.\nonumber\]This formula can also be stated as\[\int ^b_af(x)\,dx=f(c)(b-a).\nonumber\]

Proof

Since \(f(x)\) is continuous on \([a,b]\), by the Extreme Value Theorem, it assumes minimum and maximum values—\(m\) and \(M\), respectively—on \([a,b]\). Then, for all \(x\) in \([a,b]\), we have \(m \leq f(x) \leq M\). Therefore, by the Comparison Theorem, we have\[m(b-a) \leq \int ^b_af(x)\,dx \leq M(b-a).\nonumber\]Dividing by \(b-a\) gives us\[m \leq \dfrac{1}{b-a} \int ^b_af(x)\, dx \leq M.\nonumber\]Since \(\displaystyle \frac{1}{b-a} \int ^b_a f(x)\, dx\) is a number between \(m\) and \(M\), and since \(f(x)\) is continuous and assumes the values \(m\) and \(M\) over \([a,b]\), by the Intermediate Value Theorem, there is a number \(c\) over \([a,b]\) such that\[f(c)=\dfrac{1}{b-a} \int ^b_a f(x)\, dx,\nonumber\]and the proof is complete.

Interactive Element: The Average Value as a Balancing Level

The following Interactive Element might help you visualize why a continuous function must take on its average value somewhere on the interval.

Interact: Drag the orange point along the curve to raise or lower the horizontal average-value line.

Observation: The green area (curve above the line) and the red area (curve below the line) are equal exactly when the line sits at the average value \(f_{\text{avg}}\); at that height the line crosses the curve, and that crossing is the point \(c\) the theorem guarantees.

Example \(\PageIndex{1}\): Finding the Average Value of a Function

Find the average value of the function \(f(x)=8-2x\) over the interval \([0,4]\) and find \(c\) such that \(f(c)\) equals the average value of the function over \([0,4]\).

Solution

The formula states the mean value of \(f(x)\) is given by\[\displaystyle \dfrac{1}{4-0} \int ^4_0(8-2x)\,dx.\nonumber\]We can see in Figure \(\PageIndex{1}\) that the function represents a straight line and forms a right triangle bounded by the \(x\)- and \(y\)-axes. The area of the triangle is \(A=\frac{1}{2}(\text{base})(\text{height})\). We have\[A=\dfrac{1}{2}(4)(8)=16.\nonumber\]The average value is found by multiplying the area by \(1/(4-0)\). Thus, the average value of the function is\[\dfrac{1}{4}(16)=4.\nonumber\]Set the average value equal to \(f(c)\) and solve for \(c\).\[\begin{array}{rcl} 8-2c & = & 4 \\ c & = & 2 \\ \end{array}\nonumber\]At \(c=2,f(2)=4\).

Figure \(\PageIndex{1}\): By the Mean Value Theorem, the continuous function \(f(x)\) takes on its average value at \(c\) at least once over a closed interval.

Checkpoint \(\PageIndex{1}\)

Find the average value of the function \(f(x)=\frac{x}{2}\) over the interval \([0,6]\) and find c such that \(f(c)\) equals the average value of the function over \([0,6]\).

Answer

The average value is \(1.5\) and \(c=3\).

Example \(\PageIndex{2}\): Finding the Point Where a Function Takes on Its Average Value

Given \(\displaystyle \int ^3_0x^2\,dx=9\), find \(c\) such that \(f(c)\) equals the average value of \(f(x)=x^2\) over \([0,3]\).

Solution

We are looking for the value of \(c\) such that\[f(c)=\dfrac{1}{3-0} \int ^3_0x^2\,dx=\dfrac{1}{3}(9)=3.\nonumber\]Replacing \(f(c)\) with \(c^2\), we have\[\begin{array}{rcl} c^2 & = & 3 \\ c & = & \pm \sqrt{3}. \\ \end{array}\nonumber\]Since \(-\sqrt{3}\) is outside the interval, take only the positive value. Thus, \(c=\sqrt{3}\) (Figure \(\PageIndex{2}\)).

Figure \(\PageIndex{2}\): Over the interval \([0,3]\), the function \(f(x)=x^2\) takes on its average value at \(c=\sqrt{3}\).

Checkpoint \(\PageIndex{2}\)

Given \(\displaystyle \int ^3_0(2x^2-1)\,dx=15\), find \(c\) such that \(f(c)\) equals the average value of \(f(x)=2x^2-1\) over \([0,3]\).

Answer

\(c=\sqrt{3}\)

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Success in College: Cheating and Consequences

The following is an excerpt from an article at WorldNetDaily.com by Dennis Prager.

A report by Donald McCabe, professor of organization management at Rutgers University and founding president of the Center for Academic Integrity at Duke University, describes a pretty bad ethical situation in America's high schools. Seventy-five percent of students engage in "serious cheating" (such as repeated cheating and significant plagiarism).

It gets worse.

First, according to the professor, this epidemic is the culmination of three decades of increased cheating in high schools.

Second, about half of all students (not just half of all cheaters) see nothing wrong with cheating.

As one who has written and lectured on ethics for 30 years, during which time I, too, have spoken at high schools across America, I am certain that these statistics are accurate. I also believe that I know the major reasons.

One reason is that parents emphasize many things over character.

A second reason is the absurd emphasis on getting into the best college. This preoccupation on the part of parents (which often begins in preschool) has utterly distorted our children's (and high schools') priorities. Because students are taught that getting into the right college is the most important thing in their lives, they readily compromise their character to achieve this goal. And they could not care less about learning. Since the purpose of school is to get good grades, learning is incidental, and loving learning is a non-issue.

A third and final reason is the death of the concept of sin, a product of the thoroughly secular education American high schools offer. As noted, far worse than the number of students who cheat is the number of students who think that there is nothing wrong with cheating. I can testify from my talks at high schools that the vast majority of students believe what they have been taught - that right and wrong - are entirely subjective concepts. There is no morality higher than one's own; therefore, right and wrong are defined by the individual, who can rationalize any behavior.

When students do not understand a concept, they need help. Instead, some students in this situation choose to cheat. In college, any form of cheating is "academic dishonesty."

1. Has your professor discussed or posted their cheating policy?
2. What happens if you cheat in this math class?

To be clear, cheating - in any form - results in more than just a "zero" on that assignment. This includes, but is not limited to, using notes on exams, using a device that is not allowed during an exam, or using unapproved technology to help during an exam. If I suspect a student of cheating, I will have them meet me outside of class time in my office to discuss what happened and why it is considered cheating. They will receive a zero on the assignment/exam and will be reported to the Student Discipline Officer for further discussions. The Student Discipline Officer is involved for two reasons: to give the student tools to help them break the "cheating habit," and to create an official record within the Los Rios Community College District to track that student's cheating offenses - if multiple offenses have occurred, the student could like be placed on academic probation or, in severe cases, completely expelled from the District.

- Professor Simpson, Cosumnes River College

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Success in Mathematics: Holding Onto Old, Broken Techniques

It's common for students to learn a problem-solving technique years ago and not want to learn a different method because they are comfortable with their old habits. Unfortunately, we often fail to recognize that if our old, comfortable technique worked well, we should be much further along in our mathematical journey.

This adherence to comfort is widespread for all human beings in various situations. Whether it's a dead-end job, an abusive relationship, or a bad habit, we tend to ignore signals—remaining in a known situation rather than taking the chance to improve and evolve.

In general, when your professor gives you advice or a different approach to doing something, they are doing so from the lens of years of experience seeing students repeat old, broken techniques or habits that tend to lead students to make mistakes. Paying attention to these words of wisdom is critical.

EXAMPLE

Simplify the compound rational expression\[ \dfrac{-\frac{2}{(x+h)^2} + \frac{2}{x^2}}{h} \nonumber\]Students stuck in old habits tend to find the common denominator of the fractions in the numerator, make equivalent fractions, add fractions, divide the result by \(h\), then simplify. This entire process looks like the following:\[ \begin{array}{rcl} \dfrac{-\frac{2}{(x+h)^2} + \frac{2}{x^2}}{h} & = & \dfrac{-\frac{2x^2}{x^2(x+h)^2} + \frac{2(x+h)^2}{x^2(x+h)^2}}{h} \\[6pt] & = & \dfrac{\frac{-2x^2 + 2(x+h)^2}{x^2(x+h)^2}}{h} \\[6pt] & = & \dfrac{-2x^2 + 2(x+h)^2}{x^2(x+h)^2} \div h \\[6pt] & = & \dfrac{-2x^2 + 2(x+h)^2}{x^2(x+h)^2} \cdot \dfrac{1}{h} \\[6pt] & = & \dfrac{-2x^2 + 2(x+h)^2}{h x^2(x+h)^2} \\[6pt] & = & \dfrac{-2x^2 + 2(x^2+2hx + h^2)}{h x^2(x+h)^2} \\[6pt] & = & \dfrac{-2x^2 + 2x^2+4hx + 2h^2}{h x^2(x+h)^2} \\[6pt] & = & \dfrac{4hx + 2h^2}{h x^2(x+h)^2} \\[6pt] & = & \dfrac{h(4x + 2h)}{h x^2(x+h)^2} \\[6pt] & = & \dfrac{\cancel{h}(4x + 2h)}{\cancel{h} x^2(x+h)^2} \\[6pt] & = & \dfrac{(4x + 2h)}{ x^2(x+h)^2} \\[6pt] \end{array} \nonumber \]Compare this to the advice given by your math professor to begin by multiplying both numerator and denominator of the original expression by the LCD of all minor fractions within the compound rational expression.\[ \begin{array}{rcl} \dfrac{-\frac{2}{(x+h)^2} + \frac{2}{x^2}}{h} & = & \dfrac{-2x^2 + 2(x+h)^2}{hx^2(x+h)^2} \\[6pt] & = & \dfrac{-2x^2 + 2(x^2+2hx + h^2)}{h x^2(x+h)^2} \\[6pt] & = & \dfrac{-2x^2 + 2x^2+4hx + 2h^2}{h x^2(x+h)^2} \\[6pt] & = & \dfrac{4hx + 2h^2}{h x^2(x+h)^2} \\[6pt] & = & \dfrac{h(4x + 2h)}{h x^2(x+h)^2} \\[6pt] & = & \dfrac{\cancel{h}(4x + 2h)}{\cancel{h} x^2(x+h)^2} \\[6pt] & = & \dfrac{(4x + 2h)}{ x^2(x+h)^2} \\ \end{array} \nonumber \]As you can see, the second method is cleaner and less complex.

The moral of the story is, when learning, try to be open to... learn.