Why do we use integration to calculate the average of a function?

Question

What I think is that we need to sum up all the values on the curve and we can only do that by integration. Is that correct?

Answer 1

Suppose we want to get the average of $f(x)$ for $a \le x \le b$.

As a first estimate, we might use $A_2 =\dfrac{f(a)+f(b)}{2} $.

Adding another point, $A_3 =\dfrac{f(a)+f((a+b)/2)+f(b)}{3} $.

If we use $n$ points, we get $A_n =\frac1{n}\sum_{k=0}^{n-1} f(a+k(b-a)/n) $.

In the limit, if it exists, we get $A =\lim_{n \to \infty} A_n =\lim_{n \to \infty} \frac1{n}\sum_{k=0}^{n-1} f(a+k(b-a)/n) $.

If $f(x)$ is nicely behaved, $\int_a^b f(x) dx =\lim_{n \to \infty} I_n $ where $I_n =\Delta\sum_{k=0}^{n-1} f(a+k\Delta) $ and $\Delta =\Delta(a, b, n) =\dfrac{b-a}{n} $.

But, $I_n =\Delta\sum_{k=0}^{n-1} f(a+k\Delta) =\dfrac{b-a}{n}\sum_{k=0}^{n-1} f(a+k\dfrac{b-a}{n}) =(b-a)A_n $, so $A_n =\dfrac1{b-a}I_n $.

Taking the limit, the average value is the integral divided by the length of the interval.