1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I Finding the sum of heights under a curve

  1. Apr 1, 2017 #1
    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
    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?
  2. jcsd
  3. Apr 2, 2017 #2


    Staff: Mentor

    This doesn't make any sense. In any interval [a, b] of nonzero length, there are an uncountable infinity of points.

    First off, how are you going to list them? You can use x1, x2, etc., with indices from the integers, because the integers, while infinite, are only countably infinite.
    I think it doesn't make any sense for precisely the reason you say. As you evaluate the function at more and more points, you get larger and larger numbers. How is that useful?
  4. Apr 2, 2017 #3
    The limit of n to inf of the interval (b-a)/n is the step size where the function is evaluated. It's the same way a normal integral is calculated, but all I've done is remove multiplying the width by (b-a)/n AKA width of very small size.
  5. Apr 2, 2017 #4


    Staff: Mentor

    This is a crucial difference. In a Riemann integral you're taking the limit of a sum of products that could be interpreted as areas, with each of them being some small but finite width. What you're doing is just adding up more and more numbers. There's really no comparison.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted