AP Calculus AB and BC: Chapter 5 – Applications of Integration : 5.6 – Average Value of a Function Study Notes

Average Value of a Function

Definition of the Average Value of a Function
If $f$ is integrable on $[a, b]$, then the average value of $f$ on the interval is
$
\frac{1}{b-a} \int_a^b f(x) d x
$

The Mean Value Theorem for Definite Integrals
If $f$ is continuous on $[a, b]$, then there exists a number $c$ in $[a, b]$, such that

$
\int_a^b f(x) d x=f(c)(b-a)
$
In the figure when $f \geq 0$, the area of the rectangle, which is $f(c)(b-a)$, is equal to the area of the region bounded by the graph of $f$, the $x$-axis, and the vertical lines $x=a$ and $x=b$.

Example 1

  • Find the average value of $f(x)=\frac{1}{2} x \cos \left(x^2\right)+x$ on the interval $[0, \sqrt{2 \pi}]$.
▶️Answer/Explanation

Solution

Average value
$
\begin{aligned}
& =\frac{1}{\sqrt{2 \pi}-0} \int_0^{\sqrt{2 \pi}}\left(\frac{1}{2} x \cos x^2+x\right) d x \\
& =\frac{1}{\sqrt{2 \pi}}\left[\int_0^{\sqrt{2 \pi}}\left(\frac{1}{2} x \cos x^2\right) d x+\int_0^{\sqrt{2 \pi}} x d x\right] \\
& =\frac{1}{2 \sqrt{2 \pi}} \int_0^{\sqrt{2 \pi}} x \cos x^2 d x+\frac{1}{\sqrt{2 \pi}} \int_0^{\sqrt{2 \pi}} x d x
\end{aligned}
$

$=\frac{1}{4 \sqrt{2 \pi}} \int_0^{2 \pi} \cos u d u+\frac{1}{2 \sqrt{2 \pi}} \int_0^{2 \pi} d u \quad$ Let $u=x^2$, then $d u=2 x d x . \Rightarrow \frac{1}{2} d u=x d x$.

$
=\frac{1}{4 \sqrt{2 \pi}}[\sin u]_0^{2 \pi}+\frac{1}{2 \sqrt{2 \pi}}[u]_0^{2 \pi}
$
If $x=0, u=0$ and if $x=\sqrt{2 \pi}, u=2 \pi$.

$\begin{aligned} & =\frac{1}{4 \sqrt{2 \pi}}[\sin 2 \pi-\sin 0]+\frac{1}{2 \sqrt{2 \pi}}[2 \pi-0] \\ & =\frac{\pi}{\sqrt{2 \pi}} \approx 1.253\end{aligned}$

Example 2

  • Let $f$ be the function given by $f(x)=x^3-2 x+4$ on the interval $[-1,2]$. Find $c$ such that average value of $f$ on the interval is equal to $f(c)$.
▶️Answer/Explanation

Solution
Average value

$
\begin{aligned}
& =\frac{1}{b-a} \int_a^b f(x) d x=\frac{1}{2-(-1)} \int_{-1}^2\left(x^3-2 x+4\right) d x \\
& =\frac{1}{3}\left[\frac{x^4}{4}-x^2+4 x\right]_{-1}^2=4.25
\end{aligned}
$
Therefore, $f(c)=4.25$.

Use a graphing calculator to graph $y_1=x^3-2 x+4$ and $y_2=4.25$.
Find the points of intersection using a graphing calculator.
There are two points of intersection on the interval $[-1,2]$,
$(-.126,4.25)$ and $(1.473,4.25)$.
Therefore $c=-.126$ or $c=1.473$.

Example 3

  • What is the average value of $f(x)=\sqrt{x}(4-x)$ on the closed interval $[0,4]$ ?

(A) $\frac{7}{3}$

(B) $\frac{21}{5}$

(C) $\frac{32}{15}$

(D) $\frac{35}{4}$

▶️Answer/Explanation

Ans:C

Example 4

  • The graph of $y=f(x)$ consists of a semicircle and two line segments. What is the average value of $f$ on the interval $[0,8]$ ?

(A) $\frac{\pi+2}{4}$

(B) $\frac{\pi+3}{4}$

(C) $\pi+1$

(D) $\frac{\pi+6}{4}$

▶️Answer/Explanation

Ans:D

Example 5

The graph of $y=f(x)$ consists of three line segments as shown above. If the average value of $f$ on the interval $[0,5]$ is 1 what is the value of $k$ ?

(A) $\frac{3}{5}$

(B) $\frac{7}{10}$

(C) $\frac{4}{5}$

(D) $\frac{9}{10}$

▶️Answer/Explanation

Ans:C

Example 6

 

The function $f$ is continuous for $-4 \leq x \leq 4$. The graph of $f$ shown above consists of three line segments. What is the average value of $f$ on the interval $-4 \leq x \leq 4$ ?

(A) -1

(B) $-\frac{1}{2}$

(C) $\frac{1}{2}$

(D) 1

▶️Answer/Explanation

Ans:B

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