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Geometric area integration of 1/x

  1. May 31, 2009 #1

    morrobay

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    Gold Member

    given:
    ln (x) = integral 1/t dt
    from 1 to x
    and x=30

    Without a calculator and only a graph of y=1/x
    How could you show that this geometric area under this curve (with any type of unit)
    is equal to 3.4 area units, the ln (30)

    not homework, I am looking for a tangible/physical example of above definition
     
  2. jcsd
  3. May 31, 2009 #2

    HallsofIvy

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    I'm not sure what you question is. If there were a purely geometric way to show that the area under y= 1/x, from x= 1 to x= 30, we wouldn't need integral calculus!

    You could, of course, approximate it, using "Rieman sums". If we divide the x-axis, from x= 1 to x= 30, into N equal intervals, the right end of each interval is x= 1+ 29i/N with i from 1 to 30 and each interval has length 29/N. If we set up a rectangle with f(x)= f(1+ i/30)= 1/(1+ i/30)= 30/(i+30), then the area of each rectangle is (30/(i+ 30))(29/N). The area under the curve, then, is approximated by [itex]\sum 30(29)/(N(i+ 30)[/itex]. I suppose you could do that without using a calculator but I wouldn't want to! I will stick with calculus.
     
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