Calc1231W13T1 - Average of a function
Sunday, 21 October 2018
5:21 PM
Created with Microsoft OneNote 2016.
![Consider the discrete function f@) defined on [0, 8];
9
6
1
3
4
7
6
2
The sum of the values of f(x) from O to 8 is
The average value of f@) on the interval [0, 8] is:
and
and
and
and
and
and
and
and
8
6
4
2
8
n=l
f(n) —
38
4 75](Calc1231W13T1%20-%20Average%20of%20a%20function_files/image001.png)
![The average value f of a function f on [a, b] is defined by the formula
1
f@) dc.
• the average value of f@) = 5 on [2, 5] is f
• the average value of g(x) — 2m + 1 on [1, 5] is
• the average value of h@)
x +1 on [5, 10] is h
7
178/3](Calc1231W13T1%20-%20Average%20of%20a%20function_files/image003.png)
![The Mean Value Theorem for Integrals states that if f@) is continuous on the interval [a, b] then for some c e (a, b):
f(c) = j
dc.
For a fixed interval [a, b]
[2, 8] and the continuous functions f, g and h specified below; the Mean Value Theorem for Integrals guarantees that for some c e (2, 8) the function value at c
equals the function average. Using the slider below, (or otherwise) find c:
. f@) = 2m + 1 then f(c)
• Hence, c
5
= 2m + I
= 5.7
(5.7, 12.4)
Similarly if we change the interval to [a, b] = [4, 7] then the functions g, h defined below are continuous on this interval. The Mean Value Theorem for integrals then states that for some c,
the function value at c equals the function average, find c:
• g@) = 5 then c
• h(x) = 6m — 3 then c
33/6](Calc1231W13T1%20-%20Average%20of%20a%20function_files/image004.png)
