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Sums of weird series

  1. Jun 16, 2009 #1
    I have a rudimentary understanding of integration as it applies to finding the area under a curve. I get the idea of adding up the areas of progressively smaller rectangles to approach the area, and that at an infinite number of rectangles the areas would be exactly the same. Right now I'm just playing around with the idea and I'm curious about how to sum up n number of things if the ratio between each one changes.

    For example, I've drawn a graph of f(x) = [tex]\sqrt{x}[/tex] between 0 and 1. Now this isn't like a geometric series where I can find the sum using [tex]S_{n}=a_{1}(1-r^{n})/1-r[/tex], because r changes. I have discovered that [tex]a_{n} = a_{1}\sqrt{n}[/tex]. I have played around with the ratios and discovered some interesting patterns that emerge, and I have found a complicated way to sum up two objects, but it is really more work than just doing the sum directly, and there'd be no way to do it when [tex]n=\infty[/tex].

    Sorry for such a basic question, but how are things like this summed?
    Last edited: Jun 16, 2009
  2. jcsd
  3. Jun 16, 2009 #2


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    The idea of integration as the limit of Riemann sums can be used to determine what functions are integrable and can be used as a guide to setting up integrals in applications. But in fact, for all except the simplest examples, we use the "Fundamental Theorem of Calculus"-
    [tex]\int_a^b f(x) dx= F(b)- F(a)[/itex]
    where F(x) is any function having f(x) as derivative- F is an "anti-derivative" of f.

    In this particular case, to find
    [tex]\int_0^1\sqrt{x}dx= \int x^{\frac{1}{2}}dx[/tex]
    I would note that the derivative of [itex](2/3)x^{3/2}[/itex] is [itex](2/3)(3/2)x^{3/2- 1}= x^{1/2}[/itex] so I can take [itex]f(x)= x^{1/2}[/itex] and [itex]F(x)= (2/3)x^{3/2}[/itex].
    [tex]\int_0^1\sqrt{x}dx= (2/3)(1^{3/2}- 0^{3/2})= 2/3[/itex]
  4. Jun 16, 2009 #3
    Thanks for the response. I sort of understand it, except for where the derivative of [itex](2/3)x^{3/2}[/itex] comes from. I mean, is it just something that one has to work out until you find an antiderivative that equals [tex]\sqrt{x}[/tex], or is there some more basic and natural way that this comes about?
  5. Jun 16, 2009 #4


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    It is exactly for this reason that finding a primitive is a lot harder in general than finding a derivative. However in this case you can just use the power rule for integrating polynomials.

    \int x^n dx=\frac{1}{1+n} x^{n+1},\;\;n \neq -1

    In word, raise the power of your integrand by 1 then divide through the new power.
    Last edited: Jun 16, 2009
  6. Jun 16, 2009 #5


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    I guess one way you could think of it is that if you had a derivative function which measures all the changes between two points, adding up all the changes between these points gives the normal function value. In this case the derivative function is just that and the anti derivative is the function value. Its kind of intuitive if you think of it in that way and then calculus just has to make sense.
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