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Math Help - Integral Problem

  1. #1
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    Integral Problem

    Let f be a continuous function with domain x>0 and let F be the function given by F(x) = \int_1^x f(t)dt for x>0. Suppose that F(ab) = F(a) + F(b) for all a>0 and b>0 and that F'(1) = 3.

    a) Find f(1)
    b) Prove aF '(ax) = F '(x) for every positive constant a
    c) Find f(x), justify answer

    a) is easy, it is zero because going from 1 to 1 is like going nowhere.

    b) is where I am having a little difficulty with. I am not sure how to solve this. I guess I use liebniz's rule for the original integral pf F(x)?
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  2. #2
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    Quote Originally Posted by Lord Darkin View Post
    Let f be a continuous function with domain x>0 and let F be the function given by F(x) = \int_1^x f(t)dt for x>0. Suppose that F(ab) = F(a) + F(b) for all a>0 and b>0 and that F'(1) = 3.

    a) Find f(1)
    b) Prove aF '(ax) = F '(x) for every positive constant a
    c) Find f(x), justify answer

    a) is easy, it is zero because going from 1 to 1 is like going nowhere.


    I don't understand this remark: F(x) is a primitive of f(x) and thus f (1)=F'(1)=3. Perhaps you confused between f(1) and F(1)?


    b) is where I am having a little difficulty with. I am not sure how to solve this. I guess I use liebniz's rule for the original integral pf F(x)?

    You can put F(x)=\int\limits_1^xf(t)dt=G(x)-G(1), where G is a primitive of f (just as F is...), so:

    F(ab)=F(a)+F(b)\Longrightarrow G(ab)-G(1)=G(a)+G(b)-2G(1)\Longrightarrow  G(ab)=G(a)+G(b)-G(1), from where

    F(ax)=G(ax)-G(1)=G(a)+G(x)-2G(1)\Longrightarrow aF_x(ax)=G'(x)=f(x)=F'(x)

    Tonio
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  3. #3
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    Quote Originally Posted by tonio View Post
    You can put F(x)=\int\limits_1^xf(t)dt=G(x)-G(1), where G is a primitive of f (just as F is...), so:

    F(ab)=F(a)+F(b)\Longrightarrow G(ab)-G(1)=G(a)+G(b)-2G(1)\Longrightarrow  G(ab)=G(a)+G(b)-G(1), from where

    F(ax)=G(ax)-G(1)=G(a)+G(x)-2G(1)\Longrightarrow aF_x(ax)=G'(x)=f(x)=F'(x)

    Tonio
    oops, I just realized from my original post, that f(1) might not be zero since it's the capital F that has the integral from 1 to x.



    f(1) = 3 then.

    By the way, is it ok if you explain part b in words? I am a little confused as to how you got that.
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  4. #4
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    Wait, can you also explain that last line, because part b says aF'(ax)=F'(x) but you say that aF(ax)=F'(x).

    Thanks for explaining part a though.

    Does anyone else want to participate? This is due tomorrow but I can check in the morning.
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  5. #5
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    Quote Originally Posted by Lord Darkin View Post
    Wait, can you also explain that last line, because part b says aF'(ax)=F'(x) but you say that aF(ax)=F'(x).

    Thanks for explaining part a though.

    Does anyone else want to participate? This is due tomorrow but I can check in the morning.
    He wrote F_{x}(ax), which means derivative in terms of x. I can't follow all of his post either.

    He's using G(x) to be similar to the integral of f(t)dt, but more precisely G'(x)=f(x). I get stuck at this point of his post:

    F(ax)=G(a)+G(x)-2G(1)

    That I follow. But F'(ax) should equal G'(a)+G'(x)-2G'(1). G'(x)=f(x) by our definition of G and G'(1)=f(1)=3, but I don't know about G'(a), which is f(a). a*F'(ax)=aF'(a)+aF'(x) I believe. I too am curious how to put it all together. It doesn't seem to be that hard of a question but it is stumping me.
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  6. #6
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    Quote Originally Posted by Jameson View Post
    He wrote F_{x}(ax), which means derivative in terms of x. I can't follow all of his post either.

    He's using G(x) to be similar to the integral of f(t)dt, but more precisely G'(x)=f(x). I get stuck at this point of his post:

    F(ax)=G(a)+G(x)-2G(1)

    That I follow. But F'(ax) should equal G'(a)+G'(x)-2G'(1). G'(x)=f(x) by our definition of G and G'(1)=f(1)=3, but I don't know about G'(a), which is f(a). a*F'(ax)=aF'(a)+aF'(x) I believe. I too am curious how to put it all together. It doesn't seem to be that hard of a question but it is stumping me.

    When you derivate wrt x you get G'(a)=G'(1) = 0 since these are constant numbers...

    Tonio
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