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Math Help - Changing integration limits

  1. #1
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    Changing integration limits

    Hi
    Am a bit rusty on integration.

    Was wondering,

    Original integration:

    integrate 1/x between limits 1 and 2

    Problem:

    need limits to be 0 and 1. what do i change the 1/x when i change the limits?

    cheers
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  2. #2
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    Have you considered u = x-1? du = dx

    Changing the limits:
    1-1 = 0
    2-1 = 1

    \int_{1}^{2}\frac{1}{x}dx = \int_{0}^{1}\frac{1}{u+1}du
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  3. #3
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    Thanks for quick replies guys.
    TKHUNNY i think that might be right.

    Galactacus, yeah its a unusual format for a question. The aim is to apply monte carlo simulation to evaluate the integral, but this particular monte carlo simulation technique is based on evaluating the integral between 0 and 1.
    So a standard question gives you an integral with the "wrong" limits and you have to change these limits to 0 and 1 to enable monte carlo
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  4. #4
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    Hello, wjn1982!

    Original integration: . \int^2_1\frac{1}{x}\,dx

    Problem: need limits to be 0 and 1.
    How do i change the 1/x when i change the limits?

    If I understand the problem, we have a horizontal translation . . .


    The original integral represents the area under y \,=\,\frac{1}{x}\,\text{ from }x = 1\text{ to } x = 2
    Code:
                |*
                |
                | *
                |  *
                |     *.
                |     |:::::*
                |     |:::::|
          - - - + - - + - - + - -
                |     1     2

    To maintain the same area, using limit x = 0\text{ to }x = 1
    . . the graph looks like this:
    Code:
           *    |
                |
            *   |
             *  |
                *.
                |:::::*
                |:::::|
          - - - + - - + - -
                0     1

    The graph is moved one unit to the left.

    Therefore, the function would be: . y \:=\:\frac{1}{x+1}

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  5. #5
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    Great, thanks!!
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