Average Value of a Function

Author: John J Weber III, PhD Corresponding Textbook Sections:

• Section 6.5 – Average Value of a Function

Expected Educational Results

• Objective 13–1: I can compute the average value of a function on $[a,b]$$[a,b]$.
• Objective 13–2: I can use the Mean Value Theorem for Integrals to find a value $c\in [a,b]$$c\in [a,b]$ such that $f(c)=f_{\text{average}}$$f(c)=f_{\text{average}}$.

Bloom’s Taxonomy

A modern version of Bloom’s Taxonomy is included here to recognize various different levels of understanding and to encourage you to work towards higher-order understanding (those at the top of the pyramid). All Objectives, Investigations, Activities, etc. are color-coded with the level of understanding.

Mean Value Theorem [MVT] for Integrals

Theorem: Mean Value Theorem for Integrals

If $f(x)$$f(x)$ is continuous on $[a,b]$$[a,b]$, then there exists a number $c$$c$ in $[a,b]$$[a,b]$ such that

$f(c)=f_{\text{ave}}=\frac{1}{b-a}{\int }_a^{b}f(x)dx$$f(c)=f_{\text{ave}}=\frac{1}{b-a}\int_a^b{f(x)\,dx}$

that is, ${\int }_a^{b}f(x)dx=(b-a)f(c)$$\int_a^b{f(x)\,dx}=(b-a)f(c)$

Procedure: How to Use MVT for Integrals

1. Verify $f(x)$$f(x)$ is continuous on $[a,b]$$[a,b]$;
2. Find $f_{\text{ave}}$$f_{\text{ave}}$;
3. Set $f(c)=f_{\text{ave}}$$f(c)=f_{\text{ave}}$;
4. Solve for $c$$c$;
5. Identify which value(s) of $c$$c$ are in the interval $[a,b]$$[a,b]$.

Example 02: Find an exact value $c$$c$ such that $f(c)$$f(c)$ is equal to average value of $f(x)=x^2+3x+6$$f(x)=x^2+3x+6$ on $[-2,2]$$[-2,2]$.

Solution 02:

1. The polynomial, $f(x)$$f(x)$, is continuous on $(-\mathrm{\infty },\mathrm{\infty })$$\left(-\infty,\infty\right)$. So, $f(x)$$f(x)$ is continuous on $[-2,2]$$[-2,2]$.

2. From Example 01 in CPT_13a_Average_Value.html, we calculated $f_{\text{ave}}={\displaystyle \frac{22}{3}}$$f_{\text{ave}}=\dfrac{22}{3}$.

3. By MVT for Integrals, $f(c)\stackrel{\text{set}}{=}{f}_{\text{ave}}$$f(c)\stackrel{\text{set}}{=}f_{\text{ave}}$

   • By algebra, $f(c)=c^2+3c+6$$f(c)=c^2+3c+6$
   • So, ${\displaystyle c^2+3c+6\stackrel{\text{set}}{=}\frac{22}{3}}$$\displaystyle c^2+3c+6\stackrel{\text{set}}{=}\frac{22}{3}$

4. Use algebra (or Mathematica [see below]) to solve for $c$$c$:

   • $c^2+3c+6={\displaystyle \frac{22}{3}}$$c^2+3c+6=\dfrac{22}{3}$
   • $\Rightarrow 3c^2+9c-4=0$$\Rightarrow 3c^2+9c-4=0$
   • $\Rightarrow c={\displaystyle \frac{-9\pm \sqrt{9^2-4(3)(-4)}}{2(3)}}\Rightarrow c={\displaystyle \frac{-9\pm \sqrt{129}}{6}}$$\Rightarrow c=\dfrac{-9\pm\sqrt{9^2-4(3)(-4)}}{2(3)}\Rightarrow c=\dfrac{-9\pm\sqrt{129}}{6}$

5. ${\displaystyle \frac{-9-\sqrt{129}}{6}\notin [-2,2]}$$\displaystyle \frac{-9-\sqrt{129}}{6}\notin [-2,2]$. So, ${\displaystyle c=\frac{-9+\sqrt{129}}{6}\in [-2,2]}$$\displaystyle c=\frac{-9+\sqrt{129}}{6}\in [-2,2]$.

Check Your Work

Use Technology to Use Technology to Solve for c in MVT

NOTE: The use of technology in this activity is to algebraically solve for a variable. On all Assessments, you may solve for $c$$c$ without showing any work.

Mathematica

1
(* Example 02: Solve for c in f(c)=f_{avg} *)
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Solve[c^2+3c+6 == 22/3, c, Reals]

Warnings:

1. Be very careful with the syntax. Syntax is the set of rules on how to write computer code. Every software program has its own unique syntax. Some basic Mathematica syntax is located at: http://www.jjw3.com/TECH_Common_Functions.pdf.

2. Mathematica requires two $==$$==$ for the equals sign in an equation.

3. You may need parens, $($$(\,$ and $)$$\,)$, to group multiple terms in the numerator and denominator.

4. Solve[ ] has at least two arguments:

   1. The equation to be solved;
   2. The variable, i.e., $c$$c$, to be solved for;
   3. The argument $\text{Reals}$$\text{Reals}$ to force Mathematica to return only real-valued results.

5. Mathematica may return complex solutions; thus, either

   1. Identify a complex solution (the result includes the imaginary unit, $i$$i$), or
   2. Use the argument Reals.

6. For help on using the Solve[ ] function:

   1. In Mathematica, execute the code: $\text{?Solve}$$\text{?Solve}$
   2. Click on $\vee$$\or$ near the bottom-left of output
   3. Click on local
   4. Read how to use the Solve[ ] function – you will be able to copy-paste code.

7. Mathematica may return a solution to $f(c)=f_{avg}$$f(c)=f_{avg}$ that is not in $[a,b]$$[a, b]$; .these solutions should not be included in your answer.

8. Remember, correct Mathematica code will be all black except for variables.

9. To execute code (including comment codes), press and hold the SHIFT key and press the ENTER key.

Consider the embedded DESMOS graph below which diagrams the use of the Mean Value Theorem in Example 02:

The key aspects of the embedded DESMOS graph are:

1. The area under $f(x)=x^2+3x+6$$f(x)=x^2+3x+6$ on $[-2,2]$$[-2,2]$ is shown in red.
2. ${\displaystyle c=\frac{-9+\sqrt{129}}{6}}$$\displaystyle c=\frac{-9+\sqrt{129}}{6}$, represented by green vertical line is between $a=-2$$a=-2$ and $b=2$$b=2$.
3. The green vertical line [$c$$c$] and the blue horizontal line [$f_{ave}$$f_{ave}$] intersect on the red curve [$f(x)=x^2+3x+6$$f(x)=x^2+3x+6$]. In other words, $f(c)=f_{ave}$$f(c)=f_{ave}$.
4. The horizontal side of the blue rectangle is from $x=-2$$x=-2$ to $x=2$$x=2$.
5. The width of the blue rectangle is $b-a=2-(-2)=4$$b-a=2-(-2)=4$.
6. The height of the blue rectangle is ${\displaystyle f_{ave}=\frac{22}{3}}$$\displaystyle f_{ave}=\frac{22}{3}$ shown by the blue horizontal line.
7. The area of the blue rectangle is ${\displaystyle A=f_{ave}(b-a)=\left(\frac{22}{3}\right)4=\frac{88}{3}}$$\displaystyle A = f_{ave}(b-a)=\left(\frac{22}{3}\right)4=\frac{88}{3}$.
8. The red shaded area [under $f(x)$$f(x)$, i.e., ${\int }_a^{b}f(x)dx$$\int_a^b{f(x)\,dx}$] is equal to blue shaded area [${\displaystyle A=\left(f_{ave}\right)(b-a)}$$\displaystyle A = \left(f_{ave}\right)(b-a)$]; In other words, ${\int }_a^{b}f(x)dx=\left(f_{ave}\right)(b-a)\Rightarrow f_{ave}=\frac{1}{b-a}{\int }_a^{b}f(x)dx$$\int_a^b{f(x)\,dx} = \left(f_{ave}\right)(b-a) \Rightarrow f_{ave}=\frac{1}{b-a}\int_a^b{f(x)\,dx}$.

Example 03: Find an approximate value $c$$c$ [to 4 decimal places] such that $f(c)$$f(c)$ is equal to average value of $f(x)={\sin }^2(x)$$f(x)=\sin^2{\!(x)}$ on $[0,1]$$[0,1]$.

Solution 03:

1. The trigonometric function, $f(x)$$f(x)$, is continuous on $(-\mathrm{\infty },\mathrm{\infty })$$\left(-\infty,\infty\right)$. So, $f(x)$$f(x)$ is continuous on $[0,1]$$[0,1]$.

2. ${\displaystyle f_{\text{ave}}=\frac{1}{1-0}{\int }_0^1{\sin }^2(x)dx=1{\int }_0^1\frac{1}{2}(1-\cos (2x))dx={\displaystyle \frac{1}{2}}{(x-{\displaystyle \frac{\sin (2x)}{2}})}_0^1}$$\displaystyle f_{\text{ave}}=\frac{1}{1-0}\int_0^1{\sin^2{\!(x)}\,dx}=1\int_0^1{\frac{1}{2}\left(1-\cos{(2x)}\right)\,dx}=\dfrac{1}{2}\left(x-\dfrac{\sin{(2x)}}{2}\right)_0^1$ ${\displaystyle \Rightarrow f_{\text{ave}}={\displaystyle \frac{1}{2}}-{\displaystyle \frac{\sin (2)}{4}}}$$\displaystyle \Rightarrow f_{\text{ave}}=\dfrac{1}{2}-\dfrac{\sin{(2)}}{4}$

3. By MVT for Integrals, $f(c)\stackrel{\text{set}}{=}{f}_{\text{ave}}$$f(c)\stackrel{\text{set}}{=}f_{\text{ave}}$

   • By algebra, $f(c)={\sin }^2(c)$$f(c)=\sin^2{\!(c)}$
   • So, ${\displaystyle {\sin }^2(c)\stackrel{\text{set}}{=}{\displaystyle \frac{1}{2}}-{\displaystyle \frac{\sin (2)}{4}}}$$\displaystyle \sin^2{\!(c)}\stackrel{\text{set}}{=}\dfrac{1}{2}-\dfrac{\sin{(2)}}{4}$

4. Use algebra and trigonometry (and Mathematica or DESMOS) to solve for $c$$c$:

   • ${\sin }^2(c)={\displaystyle \frac{1}{2}}-\frac{\sin (2)}{4}$$\sin^2{\!(c)}=\dfrac{1}{2}-\frac{\sin{(2)}}{4}$
   • $\Rightarrow \sin (c)=\pm \sqrt{{\displaystyle \frac{1}{2}}-{\displaystyle \frac{\sin (2)}{4}}}$$\Rightarrow \sin{(c)}=\pm\sqrt{\dfrac{1}{2}-\dfrac{\sin{(2)}}{4}}$
   • $\Rightarrow c={\sin }^{-1}(\pm \sqrt{{\displaystyle \frac{1}{2}}-{\displaystyle \frac{\sin (2)}{4}}})\approx \pm 0.5494$$\Rightarrow c=\sin^{-1}{\!\left(\pm\sqrt{\dfrac{1}{2}-\dfrac{\sin{(2)}}{4}}\right)}\approx \pm 0.5494$

5. ${\displaystyle -0.5494\notin [0,1]}$$\displaystyle -0.5494\notin [0,1]$. So, ${\displaystyle c\approx 0.5494}$$\displaystyle c\approx 0.5494$.

Investigation 02

Use the MVT for Integrals to find an exact value $c$$c$ such that $f(c)=f_{\text{average}}$$f(c)=f_{\text{average}}$.

1. $y=\frac{1}{x}$$y=\frac{1}{x}$ on $[1,4]$$[1,4]$
2. $y=\sin (2x)$$y=\sin{(2x)}$ on $[0,\frac{\pi }{4}]$$[0,\frac{\pi}{4}]$
3. $y=2^x$$y=2^x$ on $[0,2]$$[0,2]$
4. $y=(x-3{)}^2$$y=(x-3)^2$ on $[2,5]$$[2,5]$
5. $y=x\cos (3x)$$y=x\cos{(3x)}$ on $[0,2\pi ]$$[0,2\pi]$
6. $y=\ln (x)$$y=\ln{(x)}$ on $[1,10]$$[1,10]$
7. $y=\frac{1}{1+x^2}$$y=\frac{1}{1+x^2}$ on $[-1,1]$$[-1,1]$
8. $y={\tan }^{-1}(x)$$y=\tan^{-1}{(x)}$ on $[0,5]$$[0,5]$

Homework

At this time, you should be able to complete the following assignments:

• Section 6.5: # 1, 3, 9, 11, 14, 17.

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Created: Thursday, 8 October 2020 7:40 EDT Last Modified: Monday, 04 July 2022 - 08:56 (EDT)