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Math Help - Mathlab integral(area)

  1. #1
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    Mathlab integral(area)

    Hello I have Mathlab installed at my computer and want to compute the integral (area) between to functions and I donīt know how and what commands to use. My two functions are (x/((11.08^0.138*4202463^0.235*5508000^0.4*1000*10^2. 806)/0.8515))^(1/-0.444)
    and x= 10


    And then is it possible to plot it?
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  2. #2
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    Quote Originally Posted by gruvan View Post
    Hello I have Mathlab installed at my computer and want to compute the integral (area) between to functions and I donīt know how and what commands to use. My two functions are (x/((11.08^0.138*4202463^0.235*5508000^0.4*1000*10^2. 806)/0.8515))^(1/-0.444)
    and x= 10


    And then is it possible to plot it?
    Why have you not simplifed your function:

    f(x)=\frac{1.37127 \times 10^{23}}{ x^{2.25225}}

    Now you need to be more precise about what area you are talking about, because you have not specified an area yet.
    (sketch the curves)

    RonL
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  3. #3
    Grand Panjandrum
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    This is the second time you have posted a question with long unsimplified products of numeric constants.

    Why is that, what is stopping you from evaluating these things?

    RonL
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  4. #4
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    Hello captain black,thanks for your help.
    I want to find the "producer surplus" (green area in the pic below).The equation (1.37127*10^23)/x^2.25225 is the demand function for driving cars and P is the price of gasoline. So how do I compute this in mathlab or any freeware program that you could suggest?
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  5. #5
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    Quote Originally Posted by gruvan View Post
    Hello captain black,thanks for your help.
    I want to find the "producer surplus" (green area in the pic below).The equation (1.37127*10^23)/x^2.25225 is the demand function for driving cars and P is the price of gasoline. So how do I compute this in mathlab or any freeware program that you could suggest?
    You could integrate the inverse function between the two prices.
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  6. #6
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    Hello I used a program called "Graph 4.3" and calulated the green area by dividing areas and my result turned out to be 8.83843E+10 .
    Can that be right?
    Last edited by gruvan; May 4th 2008 at 07:16 AM.
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  7. #7
    Grand Panjandrum
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    Quote Originally Posted by gruvan View Post
    Hello I used a program called "Graph 4.3" and calulated the green area by dividing areas and my result turned out to be 8.83843E+10 .
    Can that be right?

    Maxima gives 8.8370091\times 10^{+10} for the integral, which when we dispose of the spurious prescission is 8.837\times 10^{+10}.

    RonL
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