The Mean Value Theorem for Integrals

The Mean Value Theorem for Integrals states that a continuous function on a closed interval takes on its average value at some point in that interval. The theorem guarantees that if \(f \left(\right. x \left.\right)\) is continuous, a point c exists in an interval \(\left[\right. a , b \left]\right.\) such that the value of the function at c is equal to the average value of \(f \left(\right. x \left.\right)\) over \(\left[\right. a , b \left]\right. .\) We state this theorem mathematically with the help of the formula for the average value of a function that we presented at the end of the preceding section.

Theorem 5.3

The Mean Value Theorem for Integrals

If \(f \left(\right. x \left.\right)\) is continuous over an interval \(\left[\right. a , b \left]\right. ,\) then there is at least one point \(c \in \left[\right. a , b \left]\right.\) such that

\[f \left(\right. c \left.\right) = \frac{1}{b - a} \int_{a}^{b} f \left(\right. x \left.\right) d x .\]

This formula can also be stated as

\[\int_{a}^{b} f \left(\right. x \left.\right) d x = f \left(\right. c \left.\right) \left(\right. b - a \left.\right) .\]

Proof

Since \(f \left(\right. x \left.\right)\) is continuous on \(\left[\right. a , b \left]\right. ,\) by the extreme value theorem (see Maxima and Minima), it assumes minimum and maximum values—m and M, respectively—on \(\left[\right. a , b \left]\right. .\) Then, for all x in \(\left[\right. a , b \left]\right. ,\) we have \(m \leq f \left(\right. x \left.\right) \leq M .\) Therefore, by the comparison theorem (see The Definite Integral), we have

\[m \left(\right. b - a \left.\right) \leq \int_{a}^{b} f \left(\right. x \left.\right) d x \leq M \left(\right. b - a \left.\right) .\]

Dividing by \(b - a\) gives us

\[m \leq \frac{1}{b - a} \int_{a}^{b} f \left(\right. x \left.\right) d x \leq M .\]

Since \(\frac{1}{b - a} \int_{a}^{b} f \left(\right. x \left.\right) d x\) is a number between m and M, and since \(f \left(\right. x \left.\right)\) is continuous and assumes the values m and M over \(\left[\right. a , b \left]\right. ,\) by the Intermediate Value Theorem (see Continuity), there is a number c over \(\left[\right. a , b \left]\right.\) such that

\[f \left(\right. c \left.\right) = \frac{1}{b - a} \int_{a}^{b} f \left(\right. x \left.\right) d x ,\]

and the proof is complete.

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Example 5.15

Finding the Average Value of a Function

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

Solution

The formula states the mean value of \(f \left(\right. x \left.\right)\) is given by

\[\frac{1}{4 - 0} \int_{0}^{4} \left(\right. 8 - 2 x \left.\right) d x .\]

We can see in Figure 5.26 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 \(A = \frac{1}{2} \left(\right. \text{base} \left.\right) \left(\right. \text{height} \left.\right) .\) We have

\[A = \frac{1}{2} \left(\right. 4 \left.\right) \left(\right. 8 \left.\right) = 16 .\]

The average value is found by multiplying the area by \(1 / \left(\right. 4 - 0 \left.\right) .\) Thus, the average value of the function is

\[\frac{1}{4} \left(\right. 16 \left.\right) = 4 .\]

Set the average value equal to \(f \left(\right. c \left.\right)\) and solve for c.

\[\begin{aligned} 8 - 2 c & = & 4 \\ c & = & 2 \end{aligned}\]

At \(c = 2 , f \left(\right. 2 \left.\right) = 4 .\)

The graph of a decreasing line f(x) = 8 – 2x over [-1,4.5]. The line y=4 is drawn over [0,4], which intersects with the line at (2,4). A line is drawn down from (2,4) to the x axis and from (4,4) to the y axis. The area under y=4 is shaded.

Figure 5.26 By the Mean Value Theorem, the continuous function \(f \left(\right. x \left.\right)\) takes on its average value at c at least once over a closed interval.

Checkpoint 5.14

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

Example 5.16

Finding the Point Where a Function Takes on Its Average Value

Given \(\int_{0}^{3} x^{2} d x = 9 ,\) find c such that \(f \left(\right. c \left.\right)\) equals the average value of \(f \left(\right. x \left.\right) = x^{2}\) over \(\left[\right. 0 , 3 \left]\right. .\)

Solution

We are looking for the value of c such that

\[f \left(\right. c \left.\right) = \frac{1}{3 - 0} \int_{0}^{3} x^{2} d x = \frac{1}{3} \left(\right. 9 \left.\right) = 3 .\]

Replacing \(f \left(\right. c \left.\right)\) with c², we have

\[\begin{aligned} c^{2} & = & 3 \\ c & = & \pm \sqrt{3} . \end{aligned}\]

Since \(− \sqrt{3}\) is outside the interval, take only the positive value. Thus, \(c = \sqrt{3}\) (Figure 5.27).

A graph of the parabola f(x) = x^2 over [-2, 3]. The area under the curve and above the x axis is shaded, and the point (sqrt(3), 3) is marked.

Figure 5.27 Over the interval \(\left[\right. 0 , 3 \left]\right. ,\) the function \(f \left(\right. x \left.\right) = x^{2}\) takes on its average value at \(c = \sqrt{3} .\)

Checkpoint 5.15

Given \(\int_{0}^{3} \left(\right. 2 x^{2} - 1 \left.\right) d x = 15 ,\) find c such that \(f \left(\right. c \left.\right)\) equals the average value of \(f \left(\right. x \left.\right) = 2 x^{2} - 1\) over \(\left[\right. 0 , 3 \left]\right. .\)

The Mean Value Theorem for Integrals

Theorem 5.3