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A new limit definition of integral?

  1. Jul 3, 2014 #1
    I think i discovered a new way to define an integral, i dont know if it helps in any particular case, but its an idea worth posting i think.
    The idea is to define the height of the rectangles based on one single point of the function and then build up the next heights for the other rectangles from the dx of the function.
    Im a student of engineering in my first year, so i dont have a very rigorous math based way to show my ideas.
    I attached two pictures showing the idea and an example
    I tested it on regular functions and it works fine, the idea is to integrate a function from its derivative, maybe this could work for functions whose derivative is easier to calculate right?
     

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  2. jcsd
  3. Jul 3, 2014 #2

    Matterwave

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    At first glance, I can't really tell if this is a good definition or not. Have you tried some functions that are not linear? e^x, or 1/x or x^2?
     
  4. Jul 3, 2014 #3
    i tried with functions such as x^2 and x^3 and it works perfectly, i dont know how to attack sums involving e^k or 1/n. But the idea is solid, i tried it for definite and indefinite integral and its working. I tried to integrate f(x)=ln(x) using its derivative but i i cant get past the 1/n sum.
     
  5. Jul 3, 2014 #4

    micromass

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    Yes, it works. But it is basically the same as doing integration by parts. You have basically just written out a kind of integration by parts in its limit definition. Notice that

    [tex](b-a)f(a) + \left(\frac{b-a}{n}\right)^2 \sum_{k=0}^n (n-k)f^\prime\left(a + k\frac{b-a}{n}\right)[/tex]

    [tex] = (b-a)f(a) + b \sum_{k=0}^n f^\prime\left(a + k\frac{b-a}{n}\right) \frac{b-a}{n} - \sum_{k=0}^n \left(a + k\frac{b-a}{n}\right) f^\prime\left(a + k\frac{b-a}{n}\right) \frac{b-a}{n}[/tex]

    So by taking limits we get

    [tex](b-a)f(a) + b\int_a^b f^\prime(x)dx - \int_a^b xf^\prime(x)dx[/tex]
    [tex] = (b-a)f(a) + b(f(b) - f(a)) - \int_a^b xf^\prime(x)dx[/tex]
    [tex] = bf(b) - af(a) - \int_a^b xf^\prime(x)dx[/tex]

    Evaluating the integral by parts gives us

    [tex] = bf(b) - af(a) - [bf(b) - af(a)] + \int_a^b f(x)dx[/tex]
    [tex] =\int_a^b f(x)dx[/tex]

    which is exactly what we want.
     
    Last edited: Jul 3, 2014
  6. Jul 4, 2014 #5

    Matterwave

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    And that's why Micro gets paid the big bucks...

    I was originally guessing that somehow you were using a Taylor approximation for a function to find the integral of said function, which would work for analytic functions, but I was getting stuck on the fact that no higher order derivatives appear.
     
  7. Jul 4, 2014 #6
    Well, for someone that has only seen first semester of calculus class i guess that big deal for me!.
    I have seen integration by parts, but i never made the connection. Makes a lot of sense i guess.
    I was actually trying to find a way to define the integral as a limit not involving sums of any kind, and then this came up to my mind.
    By the way, is there any possibility of defining an integral not involving sums? like the definition of derivative? I hate how complicated some integrals get to be
     
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