2.8: The Mean Value Theorem for Integrals

Note to the Instructor (click to expand) This is an incredibly short section - perhaps a half lecture. To succeed in this section, you'll need to use some skills from previous courses. While you should already know them, this is the first time they've been required. You can review these skills in CRC's Corequisite Codex. If you have a support class, it might cover some, but not all, of these topics. Prerequisite Skills and Support Topics (click to expand) Solving Equations 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.

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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.

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 \(A=\frac{1}{2}(base)(height).\) 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

\(c=\sqrt{3}\)