Apostol Vol.1 Chap. 2 execises 2.17 n°15

Question

i was trying to solve exercise 15 of the list of exercises 2.17 of Apostol Calculus Vol.1, but i am having difficulty finding a way to prove what he asks, that is :

$``$ Let $A_a^b(f)$ denote the average of $f$ on an interval $[a, b]$. (a) If $a < c < b$, prove that there is a number $t$ satisfying $0 < t < 1$ such that $A_a^b(f) = t A_a^c(f) + (1-t)A_c^b(f).$

Thus, $A_a^b(f)$ is a weighted arithmetic mean of $A_a^c(f)$ and $ A_{c}^{b}(f) $. $"$

I already know that choosing a convenient value for $t$ will solve this exercise, but i still wanted to know if there is any way in wich i can obtain this value instead of just pluging it in and observing that it solves the problem.

Answer 1

You might reason like this: the interval $[a,c]$ accounts for a proportion $\frac{c - a}{b - a}$ of the length of the interval $[a,b]$. Likewise, $[c,b]$ accounts for a $\frac{b - c}{b - a}$ proportion of the length of $[a,b]$. The average $A_a^b(f)$ is the weighted average of $A_a^c(f)$ and $A_c^b(f)$ with the same proportions.

Answer 2

The simple way to do this is to show that for $t = 0$ you get one thing, and for $t = 1$ you get another, and therefore for $t = $ something, you have to get the desired value, i.e., apply the intermediate value theorem. (Of course, there are details to work out). You need to know, for instance, that the thing you're working with is a continuous function of $t$. And you need to know that the value you're looking for is between the end-point values.

The second way to do it is to figure out what $t$ must be (i.e., which linear combination of the two end-values works), and then show that $t$ must be between $0$ and $1$ so that you actually get an affine combination (or "weighted arithmetic mean") rather than a general linear combination like $2.5 v_1 + (1-2.5) v_2$.

It's not clear to me that either of these is better than the other.

Answer 3

You could break it into cases. In the case where the average values are equal, then any $t$ satisfies the equation. Note that equality of any two of the averages implies equality of the third.

In the case where they are unequal, you obtain an expression for $t$ that you can reason algebraically (and with basic integral properties) is bounded between 0 and 1.

Case 1: $A_a^b(f) = A_a^c(f) = A_c^b(f)$

Case 2: $t = \frac{A_a^b(f) - A_c^b(f)}{A_a^c(f) - A_c^b(f)}$