# The relationship of arithmetic mean and integration

The arithmetic mean of $y_i\ldots y_n$ is: ${1\over n}\sum_{i=1}^n y_i,$

If $n$ goes infinty, how to prove $\lim_{n\rightarrow \infty}{1\over n}\sum_{i=1}^n y_i={\int_{x_0}^{x_1} f(x) dx \over x_1-x_0},$

where $f(x)$ is a smooth function, $f(x_i)=y_i$ and $x_0<\ldots<x_i<\ldots<x_n<\ldots<x_1.$

• The Trapezoidal Rule? – Ray Nov 8 '17 at 22:39
• Hi, I may have interpreted your question incorrectly, but to find the average value (arithmetic mean) of a function over an interval it is $\frac{\int_a^b f(x)dx}{b-a}$, a derivation can be found here;tutorial.math.lamar.edu/Classes/CalcI/… – frog1944 Nov 8 '17 at 22:46
• Thanks frog, good answer – Ray Nov 8 '17 at 22:53

Hint: The notation is bad because $x_1$ has two meanings (it can be confused with one of the $x_i$).
But forgetting about that, note that if you multiply both sides by $x_1-x_0$, you have the sum on the left side as $$\sum_{i=1}^n y_i\;\Delta x$$ where $\Delta x\equiv\frac{x_1-x_0}{n}$.
• thanks, does it mean $f(x)$ must be a smooth function? – Ray Nov 8 '17 at 22:54