1.8: The Mean Value Theorem for Integrals

Learning Objectives

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

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

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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 analogue 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 we can 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\[ m \leq \frac{1}{b−a} \int ^b_af(x)\,dx \leq M. \nonumber \]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\[ f(c)=\frac{1}{b−a} \int ^b_a f(x)\,dx, \nonumber \]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.

Exercise \(\PageIndex{1}\)

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 \([0,6].\)

Hint

Use the procedures from Example \(\PageIndex{1}\) to solve the problem

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

\[f(c)=\frac{1}{3−0} \int ^3_0x^2\,dx=\frac{1}{3}(9)=3. \nonumber \]

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 \(c=\sqrt{3}\).

Exercise \(\PageIndex{2}\)

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 \([0,3]\).

Hint

Use the procedures from Example \(\PageIndex{2}\) to solve the problem.

Answer

\(c=\sqrt{3}\)

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Key Concepts

• The Mean Value Theorem for Integrals states that for a continuous function over a closed interval, there is a value c such that \(f(c)\) equals the average value of the function.

Key Equations

• 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 \[f(c)=\frac{1}{b−a} \int ^b_af(x)\,dx.\nonumber \]

Glossary

Mean Value Theorem for integrals

guarantees that a point \(c\) exists such that \(f(c)\) is equal to the average value of the function