calculus average value

[bobobob111]

We are given a function, g(x), with the following properties
- the function, g(x), is continuous
- the average value of f over the interval [0, 1] = 1
- the average value of f over the interval [1, 3] = 2
- the average value of f over the interval [3, 6] = 3
By using the definition of average value, find the average value of f over [0, 6].

I've tried just adding all 3 average values of the subinterval values and I got 6 (1+2+3). Then I tried multiplying that by 1/b-a of the entire interval (1/6=0=1/6) and got 1 for my final answer. Can you not just add the average value of the intervals like that to find the average value of the whole interval? I'm very confused.

 

[Cubist]

It might help you out if you state the definition of average value, and then think about the following question:-

If the average value of f over the interval [3, 6] = 3, then what is the integral of f over the interval [3, 6]?

 

[Steven G]

Adding to Cubist excellent response. Draw a graph of the average value over each given interval

 

[HallsofIvy]

Did nobody else notice that we are told that "g(x) is continuous" but then are asked about "f(x)"?

 

[Steven G]

Did nobody else notice that we are told that "g(x) is continuous" but then are asked about "f(x)"?

Nope, but I do now.

 

[HallsofIvy]

The average value of f(x) over interval [a, b] is \(\displaystyle \frac{\int_a^b f(x)dx}{b- a}\).
Since
"- the average value of f over the interval [0, 1] = 1"
\(\displaystyle \frac{\int_0^1 f(x)dx}{1- 0}= 1\) so \(\displaystyle \int_0^1 f(x)= 1\)

"- the average value of f over the interval [1, 3] = 2"
\(\displaystyle \frac{\int_1^3 f(x)dx}{3- 1}= 2\) so \(\displaystyle \int_1^3 f(x)= 4\)

"- the average value of f over the interval [3, 6] = 3"
\(\displaystyle \frac{\int_3^6 f(x)dx}{6- 3}= 3\) so \(\displaystyle \int_3^6 f(x)dx= 9\)

Since \(\displaystyle \int_0^6 f(x)dx= \int_0^1 f(x)dx+ \int_1^3 f(x)dx+ \int_3^6 f(x)dx\)
\(\displaystyle \int_0^6 f(x)dx= 1+ 4+ 9= 14\).

So the average value of f(x) over the interval [0, 6] is \(\displaystyle \frac{14}{6}= \frac{7}{3}\).

That could also be done as a "weighted average", weighting each of the separate averages by the length of the interval: \(\displaystyle \frac{1(1)+ 2(2)+ 3(3)}{1+ 2+ 3}= \frac{14}{6}= \frac{7}{3}\).