MATH 136
Average Value and
The Mean Value Theorem for Integrals
Let f be a continuous function on the interval [ a , b ]. By the Fundamental Theorem of
Calculus, we know that the (signed) area between the graph of f and the x -axis on the
b
interval [ a , b ] is given by Area( f , [a, b]) = ∫ f (x) dx .
a
€ €
€
€ €
Ave Ht × Length gives same area
Area( f , [a, b])
However, geometrically, this area also can be expressed as
Area( f , [a, b]) = Average Height × Length = Average Height × (b − a) .
Therefore,
b
Average Height × (b − a) =
∫ f (x) dx .
a
Thus we can solve for the average height, or more generally the average value for the
case when the function is not always positive.
Definition. The average value of the function f over the interval [ a , b ] is given by
1
Avg f =
b−a
b
∫ f (x) dx .
a
€ €
80
Example 1. Let f (x) = 3 . Find the average value of f over the interval [2, 5].
x
On the interval [2, 5], the function values range from 0.64
(when x = 5) to 10 (when x = 2). Thus, the average value must
be somewhere between 0.64 and 10. (Note that the middle of
this range is (0.64 + 10)/2 = 5.32.
€
€
Dividing the integral by the length (5 – 2) = 3, the average value is given by
1
Avg f =
3
5
5
−40 −2 5 −40  −40 
80
1
−3
x
dx
80x
dx
=
=
=
−
= 2.8.
∫ x3
3
3∫
75  12 
2
2
2
Because the graph decreases so quickly, the average value of 2.8 is much nearer to
0.64 than it is to 10 and it is below the middle range value of 5.32.
The fnInt Command
To avoid integrating “by hand,” you can use the built-in integrate command on your
calculator.
On TI-84: First enter the function into Y1 and then Quit to return to the Home screen.
Press MATH and scroll down to fnInt (or press 9). The command appears on the
Home screen. To find the average value on the interval [2, 5], enter fnInt(Y1, X, 2, 5)/3.
On TI-89: First enter the function into y1 and then quit to return to the Home screen.
Press F3 for CALC and then press 2 for the integrate command. The command ∫ (
appears on the Home screen. To compute the average value on the interval [2, 5], enter
∫ (y1(x), x, 2, 5)/3.
Other Average Values
Average values often occur in more common forms involving discrete functions; that is,
non-continuous functions having only a finite number of values in the range. No
integration (nor any calculus) is needed to find the average value in these cases.
Points Per Game: In the first eight games of the season, a basketball player scores 12,
14, 22, 9, 15, 8, 13, and 16 points. What is the scoring average so far?
Avg =
12 + 14 + 22 + 9 + 15 + 8 + 13 + 16
= 13.625 ≈ 13.6 ppg.
8
Notice that the average is about 13.6 point per game, but that it is impossible to ever
score exactly 13.6 points in a game.
GPA: Here is a student’s report card from
Fall 2010. Find the grade point average.
Each grade is assigned a point value; so the
grades are 4.0, 4.0, 3.0, 2.0, 3.0, 4.0. But we do
not simply add these numbers and divide by 6.
Course
UC 101
ENG 100
HIST 119
BIOL 120
BIOL 121
MATH 136
Hrs
2.0
3.0
3.0
3.0
1.0
4.0
Gr
A
A
B
C
B
A
We must weight the grades appropriately according to credit hours. The total
number of hours taken was 16, so the GPA is

2  
3 
3 
3 
1 
4
 4.0 ×
 +  4. 0 ×  +  3. 0 ×  +  2.0 ×  +  3. 0 ×  +  4.0 ×  =











16.
16
16
16
16
16 
(4 × 2) + (4 × 3) + (3 × 3) + (2 × 3) + (3 × 1) + (4 × 4)
= 3.375.
16
 2 if 0 ≤ x < 20

 5 if 20 ≤ x < 50
Step Function: Let f ( x) = 
. Find the average value of f .
16 if 50 ≤ x < 60

20 if 60 ≤ x ≤ 100
On the interval [0, 100], the function is 2 for 20% of the time. Then the function is 5
for 30% of the time. Then the function is 16 for 10% of the time, then it is 20 for 40% of
the time. Thus the average value is
(2 × 0.20) + (5 × 0. 30) + (16 × 0.10) + (20 × 0. 40) = 11.5.
Again note that even though the average is 11.5, the function itself never equals 11.5
The Mean Value Theorem for Integrals
Unlike the discrete functions in the previous examples, we now let f be a continuous
function on the interval [ a , b ]. The Mean Value Theorem for Integrals states that there
must be a point x in the interval [ a , b ] where the exact function value actually equals
the average function value. That is, there is a solution to the equation
€ €
€
€ €
1
f (x) =
b−a
b
∫ f (x) dx .
a
We can find the point x several ways: (i) Find the average value c “by hand” and
then solve f (x) = c “by hand.” (ii) Find the average value c using the fnInt command
and then solve f (x) = c using the solve command. (iii) Use a program that computes
and displays both the average value c and the solution to f (x) = c .
€
€
€
€
80
Example 2. Let
€ f (x) = 3 on the interval [2, 5]. Find the point guaranteed by the Mean
x €
€
Value Theorem for integrals.
Solution. In Example 1, we found that the average function value
1
is Avg f =
3
5
80
3 dx = 2.8. Since the function is continuous, there
x
2
is a point in the interval [2, 5] where the function equals 2.8.
∫
1
80
80
 80  3
So now we must solve the equation 3 = 2.8 , or
= x 3 , which gives x = 
2.8 
2.8
x
≈ 3.0571. Thus when x ≈ 3.0571, then f (x) equals its average value of 2.8
€
The Solve command
To avoid solving “by hand,” you can use the built-in solve command. A good initial
guess to use is the midpoint of the interval [ a , b ].
€ €
On TI-84: First enter the function in Y1 and then Quit to return to the Home screen.
Press CATALOG (2nd 0) and press 4 to get to the T's. Scroll up to solve( and press
ENTER. The command appears on the Home screen. To solve f (x) = c , enter
solve(Y1–c, X, g) where g is the initial guess.
On TI-89: First enter the function in y1 and then quit to return to the Home screen.
Press F2 for Algebra and then press 1 for the solve command. To solve f (x) = c , enter
solve(y1(x) = c, x = g) where g is the initial guess.
2
Example 3. Let f (x) = 9x − x on [2, 8]. Find the average value of f and the point
guaranteed by the Mean Value Theorem for Integrals.
8
1  9 2 1 3 8
1
2
 x − x 
(9x
−
x
)
dx
Solution. First, Avg f =
=
= 17.
62
3  2
8−2 ∫
2
So the average value is 17. From the graph, we see that there are actually two
places in the interval [2, 8] where the function equals 17. To find them, we must solve
2
2
the equation 9x − x = 17, or 0 = x − 9x + 17 . Using the quadratic formula, we obtain
9 ± 13
the solutions of x =
, or x ≈ 6.30278 and x ≈ 2.69722.
2
We also can obtain these numerical values with the solve( command:
€
€
The programs below will compute the average value and find one of the points
guaranteed by the MVT for Integrals. To execute the program, the desired function
must first be entered into Y1.
For TI-84
PROGRAM:MVTI
:Disp "A"
:Input A
:Disp "B"
:Input B
:fnInt(Y1,X,A,B)/(B–A) → C
:Disp "AVE VALUE",C
:solve(Y1–C,X,(A+B)/2) → D
:Disp "MVTI PT",D
€
€
For TI-89
mvti()
:Prgm
:Disp "a"
:Input a
:Disp "b"
:Input b
:nInt(y1(x),x,a,b)/(b–a) → c
:Disp "ave value",c
:solve(y1(x)=c,x=(a+b)/2) → d
:Disp "mvti pt",d
€
:EndPrgm
€