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  1. #1
    Newbie nigromante's Avatar
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    defined integral

    How to demostrate that Integrate of |t|dt in {0,x} for any x in R is (1/2)x|x|
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    MHF Contributor MarkFL's Avatar
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    Re: defined integral

    Hint: use the derivative form of the fundamental theorem of calculus.
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  3. #3
    Newbie nigromante's Avatar
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    Re: defined integral

    i dont know how
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    MHF Contributor MarkFL's Avatar
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    Re: defined integral

    We are given to verify:

    (1) \int_0^x |t|\,dt=\frac{1}{2}x|x|

    The derivative form of the fundamental theorem of calculus is:

    \frac{d}{dx}\int_a^x f(t)\,dx=f(x)

    Use this for the left side, and on the right use the product rule and the fact that we have 0\le x.
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  5. #5
    Newbie nigromante's Avatar
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    Re: defined integral

    I have tried, but I can't see the relation, excuseme
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  6. #6
    Newbie nigromante's Avatar
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    Re: defined integral

    first I wish to use that t^2/2 is equal to |t|, but im not shure, that it is correct, and in the form that you sugest, I think is better, but sorry, I know that I need more practice, or I have a problem in understand a detail.
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  7. #7
    MHF Contributor MarkFL's Avatar
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    Re: defined integral

    Let's focus on the left side first. What do you get using the theorem I suggested when you differentiate?
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  8. #8
    Newbie nigromante's Avatar
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    Re: defined integral

    |t|, {0,x}=f(x)
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  9. #9
    Newbie nigromante's Avatar
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    Re: defined integral

    Or i don´t know if must be only |x|=f(x)
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  10. #10
    MHF Contributor MarkFL's Avatar
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    Re: defined integral

    Yes, you get |x|. Now, what do you get when you differentiate the right side?

    edit: Use the definition |x|\equiv\sqrt{x^2}.
    Last edited by MarkFL; November 3rd 2012 at 09:44 PM.
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  11. #11
    Newbie nigromante's Avatar
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    Re: defined integral

    F'(x)=[(x^2)/2|x|]+(|x|/2)
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  12. #12
    MHF Contributor MarkFL's Avatar
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    Re: defined integral

    On the right side, you have:

    \frac{1}{2}x|x|=\frac{1}{2}x\sqrt{x^2}

    Using the product, power and chain rules, we find:

    \frac{d}{dx}\left(\frac{1}{2}x\sqrt{x^2} \right)=\frac{1}{2}\left(x\frac{x}{\sqrt{x^2}}+ \sqrt{x^2} \right)=\frac{x^2}{\sqrt{x^2}}=\sqrt{x^2}=|x|

    Thus, we have shown the given result for the definite integral is valid for all real x.
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  13. #13
    Newbie nigromante's Avatar
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    Re: defined integral

    OOOOOOh!!!!

    It's realy easy, I was near but I didn't know, when I had to use |x| or x^2/2.

    I realy thank you for your patience and your time. . .
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  14. #14
    Newbie nigromante's Avatar
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    Re: defined integral

    I only sitll have a doubt, what about the condition that x < or = 0?
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  15. #15
    MHF Contributor MarkFL's Avatar
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    Re: defined integral

    That condition was an error on my part. x can be any real number, as originally stated in your first post.
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