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Math Help - exponential functipn definite integral

  1. #1
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    exponential function definite integral

    I have a function:

    f(x) = 10.767(1.346)^x

    I have found the area below the graph between x = -8, and x= 0, below y = 8. This comes out to 31.557.

    I need to change my function f(x) so that the area between x -8, x = 0, below y = 8 is equal to different areas.

    The areas I am testing are:

    1.25 (31.557)
    1.5(31.557)
    2(31.557)

    How would I go about doing this?, It's been bugging me.
    Last edited by Mr.Ree; August 9th 2009 at 11:32 PM. Reason: can't spell !
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  2. #2
    Member eXist's Avatar
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    What exactly are you changing in f(x)?
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  3. #3
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    a and b
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  4. #4
    Member eXist's Avatar
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    f(x) = 10.767(1.346)^x

    I don't see an a or b in that equation. Are you talking about the intervale from -8 to 0?
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  5. #5
    Senior Member DeMath's Avatar
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    Quote Originally Posted by Mr.Ree View Post
    I have a function:

    f(x) = 10.767(1.346)^x

    I have found the area below the graph between x = -8, and x= 0, below y = 8. This comes out to 31.557.

    I need to change my function f(x) so that the area between x -8, x = 0, below y = 8 is equal to different areas.

    The areas I am testing are:

    1.25 (31.557)
    1.5(31.557)
    2(31.557)

    How would I go about doing this?, It's been bugging me.
    Do you need this

    10.767{\left( {1.346} \right)^x} = 8 \Leftrightarrow {\left( {1.346}\right)^x} = \frac{8}{{10.767}} \Leftrightarrow x = {\log _{1.346}}\frac{8}{{10.767}} \approx  - 1.

    A = \int\limits_{ - 8}^{ - 1} {10.767{{\left( {1.346} \right)}^x}dx}  + \int\limits_{ - 1}^0 {8dx}  = \left. {\frac{{10.767}}{{\ln 1.346}}{{\left( {1.346} \right)}^x}} \right|_{ - 8}^{ - 1} + \left. {8x} \right|_{ - 1}^0 \approx 31.55769686
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