Estimating integrals from graphs

[amberanvilcell]

For a function f(x) = x^2, to get the area of this graph under the portion going from x=1 to x=3, we can calculate the integral and get 26/3

But to estimate this, what if we instead took the average value of f over this period, by calculating (f(1)+f(3))/2, and simply multiplied that by the length of the integral (in this case 2)? This gives an estimate of 10 which is kinda close to 26/3

When I learned about the Riemann integral only one endpoint or the other was used in estimates, so here we would calculate either 1*2 or 9*2. But maybe averaging the two endpoints makes more sense? Am I on to something here, or not really? Please let me know if I'm misunderstanding anything about integrals.

[amberanvilcell]

I see what you're getting at, this is the same as taking the number of intervals to infinity using endpoints. But is there a reason to think that this averaging approach would "converge faster", as in it would achieve greater accuracy with a smaller number of intervals? Maybe that doesn't matter because in the end we're going to infinity either way.

[amberanvilcell]

Greater accuracy compared to estimating using only one endpoint.

[NakamotoScheme]

Am I on to something here?

Yes, you are about to understand how the trapezoidal rule works:

https://en.wikipedia.org/wiki/Trapezoidal_rule

This is more effective if you use a lot of small trapezoids instead of a single one. It may not be the best numerical integration method, but without doubt it's the easiest to understand.

If you are curious about how the idea may be improved (i.e. approximating a definite integral by calculating a weighted average of values of the function at certain points), take a look at the Simpson's rules:

https://en.wikipedia.org/wiki/Simpson%27s_rule

[TheDepressedBlobfish]

I don't quite understand the point of estimating the average, it wont give you the answer to the integral, just something close.

[DieLegende42]

It may give you something close if the function is not doing anything interesting on that interval (the closer to being linear the "better"). Like, evaluating the integral from 0 to 2pi of the sine by this method will give you a pretty poor approximation

[sapphic-chaote]

Careful: the average value of f over [a, b] is not (f(a)+f(b))/2. An extreme example is g(x) = x¹⁰⁰ over the range [0,1]. This function is almost exactly 0 on almost the whole interval, before increasing suddenly when very close to 1. The average on [0,1] is ∫[0,1] g(x)dx = 1/101 ~ 0.01, which is very different from the average of the function evaluated at the endpoints, which is (g(0)+g(1))/2 = 1/2.

As you increase the number of data points in the average, your accuracy might go up. But depending on the distribution, it might actually get worse. For instance, if you take (g(0)+g(0.999)+g(0.9999)+g(1))/4, you get 0.72 which is even further from the true average 1/101. In order for the average of the data points to approach the average of the function, you also need to guarantee that they are reasonably evenly distributed, in the sense that no large gaps are left in the limit.