actual meaning of average - and implying it to functions

Question

The very basic idea of average is:

$$ average=\frac{{x_1} + {x_2}....{x_n}}{n}\ $$

A function is nothing but just a graphical representation of a set of points(in the cartesian plane in this instance):

now on the interval [a, b], the the average of the function should had been

$$average = \frac{{y_1} + {y_2}......{y_n}}{N}$$, where n is supposed to be the number of y in the interval. Instead, y_av is treated as the area divided by it's interval b-a.

Now, the average's definition could also say, finding one such value which could approximate to a certain range it's entire values of concern:

Again, interpreting a function's average as an area would give us the value of on square of a unit which would be a centrally approximated value of it's entire area; but, what I'm confused at is that y1, y2, etc.. are offset from the x axis and not functions themselves representing the entire range of points. So, how would the definition of average and the interpretation of area towards the average of a function make sense?

Answer 1

For a function $f : \mathbb R \to \mathbb R$, the average of the function on an interval $[a,b]$ is given by

$$\frac{1}{b-a}\int_a^b f(x)\,dx = \lim_{n\to\infty} \frac{1}{n} \sum_{i=1}^n f\left(a + i\cdot\frac{b-a}{n}\right) = \lim_{n\to\infty} \frac{y_1 + y_2 + \cdots + y_n}{n}$$

where the $y_i$ are the function values at points in the interval $[a,b]$ given by $y_i = f\left(a + i\cdot\frac{b-a}{n}\right)$.

The integral is the area under the curve and we divide by $b-a$, which is the length of the interval.

Answer 2

Imagine we want to find the are under a graph called $f$. If we know that the average of $f(x)$ starting from $x = a$ to $x = b$ is equal to $n$, we can say that the area under the curve is:

$n * (b-a)$

So in order to calculate n, we can do this:

$\int_{a}^{b}{f(x).dx} = n * (b-a) => n = \frac{\int_{a}^{b}{f(x).dx}}{b - a}$

Which is the same as "the area under the curve / $\Delta{x}$".