In integral calc, you add up very small areas to find the total area under the curve. So it would be f(x1)Δx + f(x2)Δx+ ..., summed up. But what if you wanted to find out the sum of all heights under the curve? So it would be something like f(x1) + f(x2) + .... I'm thinking the formulation would be this: Take bounds [a,b], and divide that into many infinite n parts to get a step size of (b-a)/n, lim n -> inf. Using right end points, evaluate f(x) where x = a + i*(b-a)/n, and i goes from 1 to inf So it then becomes Σi=1i=n f(xi) For f(x) = 2x, from 0 to 5 5-0/n, lim n ->inf xi = 0 + 5i/n = 5i/n Σi=1i=n 2xi and i = n at the end since you need to add up as many segments as you made (n segments) Σi=1i=n 2(5i/n) (10/n)Σi=1i=n i (10/n)(1+2+3+4+5...+n) and since (1+2+3+4+5...+n) is greater than n by an infinite amount then (10/n)(1+2+3+4+5...+n) = infinity? I think this result makes sense in that there are infinitely many x points to choose under the curve, and they all have a positive height (except for at x=0), so infinitely many positive heights added up = infinity. What do you think?