While playing with numerical integration I arrived at a general formula for
a numerical integral as follows.

Start with the simple function such as y = x^2 and integrate it from 0 to 5
in 5 steps.
At each step the area between the curve and the x axis is approximated by a
rectangle.
This will obviously give an over estimate.
I can write it as:

(5/5)*(1/((5/5)^2))*((1^2)+(2^2)+(3^2)+(4^2)+(5^2) = 55 which is higher than
the exact answer of 41+2/3 as expected.

If I do the same with 10 steps I can write it as
(5/10)*(1/((10/5)^2))*(1^2+2^2+3^2+4^2+5^2+6^2+7^2+8^2+9^2+10^2) = 48.125 a
bit closer

And for 20 steps I have

(5/20)*(1/((20/5)^2))*(1^2+2^2+3^2+4^2+5^2+6^2+7^2+8^2+9^2+10^2+11^2+12^2+13^2+14^2+15^2+16^2+17^2+18^2+19^2+20^2
= 44.84375

For 1000 steps I can put the following in Wolfram and get 41.7292
(5/1000)*(1/(1000/5)^2) * sum from n = 1 to 1000 of n^2

For n steps I have (5/n)*(1/((n/5)^2)) * the sum of the squares from 1 to n

More generally integrating x^2 from 0 to X instead of 0 to 5 I have, for n
steps

(X/n)*(1/((n/X)^2)) * the sum of the squares from 1 to n

Even more generally for f(x) instead of x^2 it seems to be the case that the
integral is

(X/n)*(1/(f(n/X)) * the sum of f(x) for x = 1 to n

n must be an integer but if I set n = X * k I can choose k to make n an
interger.

Then I have the following general formula for the integral of f(x) from 0 to
X

Integral from 0 to X of f(x) is approximately (1/(k*f(k)) * the sum of f(x)
for x = 1 to n

Where n = X*k so first choose the number of steps n and then k = n/X

So if I want the integral from 0 to 4 of x^3 (which is exactly 64) I could
do it in 1000 steps by putting this in Wolfram
(1/((1000/4)*((1000/4)^3)) * the sum of x^3 for x = 1 to 1000

My guess is that this method can't be generalized to functions other than
x^n or can it?

I've never seen it used anywhere else, probably because it's not of great
interest.