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How to calculate the integral as a limit of sum?

  1. Jun 22, 2012 #1
    1. The problem statement, all variables and given/known data

    The problem is to find the integral of function [itex]\frac{1}{x}[/itex] using the definition i.e the area under the curve as a limit of a sum?

    2. Relevant equations
    [itex]\int_a^b \frac{1}{x} dx = ln(\frac{b}{a})[/itex]

    3. The attempt at a solution
    I tried with dividing the interval [a,b] into n parts such that [itex] h=\frac{b-a}{n}[/itex] but what I got was a too vague expression of the sum.

    [itex] \lim_{n \rightarrow ∞} \frac{b-a}{n} = h[/itex]
    [itex]\int_a^b \frac{1}{x} dx = \lim_{h \rightarrow 0}\bigg[ \frac{1}{a}×h+\frac{1}{a+h}×h+\frac{1}{a+2h}×h^2...\bigg][/itex]

    how do I proceed now
  2. jcsd
  3. Jun 22, 2012 #2
    So is the idea that you should end up with the power series for ln(x)?
  4. Jun 23, 2012 #3
    Something like that; probably some limit of [itex]ln(x)[/itex] because [itex]\lim_{h\rightarrow 0} [/itex] is there intact throughout.
    Last edited: Jun 23, 2012
  5. Jun 23, 2012 #4


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    Maybe that's a typo, but that h should not be squared in your last line.

    [itex]\int_a^b \frac{1}{x} dx = \lim_{h \rightarrow 0}\bigg[ \frac{1}{a}×h+\frac{1}{a+h}×h+\frac{1}{a+2h}×h^2...\bigg][/itex]

    should be

    [itex]\int_a^b \frac{1}{x} dx = \lim_{h \rightarrow 0}\bigg[ \frac{1}{a}×h+\frac{1}{a+h}×h+\frac{1}{a+2h}×h ...\bigg][/itex]
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