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Math Help - Another integration problems

  1. #1
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    Another integration problems

    Ok, I solved it.
    I got f(0) = 3. Is that right?


    I then have another problem:

    f(x) = definite integral from 0 to x : sin(t) / (t+1) dt.

    1) Find where f(x) has local maximum and minimum values (I should not find the function values).

    A more interesting one:
    2) Show that f(x) > 0, when x > 0. (Here I should use integration by parts...)

    Do anyone want to help? :-)
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  2. #2
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    Quote Originally Posted by expresstrain View Post
    Ok, I solved it.
    I got f(0) = 3. Is that right?
    i split your topic into two ones.

    don't know if that is the correct answer, i have no time to do the problem.



    Quote Originally Posted by expresstrain View Post
    f(x) = definite integral from 0 to x : sin(t) / (t+1) dt.

    2) Show that f(x) > 0, when x > 0. (Here I should use integration by parts...)

    Do anyone want to help? :-)
    Let \displaystyle\int_0^x g(t)\sin(t)\,dt where g(t) is class 1, positive and decreasing for [0,\infty), then integrate by parts and get \displaystyle\int_{0}^{x}{g(t)\sin (t)\,dt}=g(x)(1-\cos x)-\int_{0}^{x}{g'(t)(1-\cos t)\,dt}.

    since g(x)>0, g'(t)<0 and 1-\cos t\ge0, the whole RHS is positive and we're done.

    try to do part a) by yourself it's just a matter of FTC application.
    Last edited by Krizalid; November 10th 2010 at 10:25 AM.
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  3. #3
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    Quote Originally Posted by expresstrain View Post
    Ok, I solved it.
    I got f(0) = 3. Is that right?
    I have no idea! What was the question?


    I then have another problem:

    f(x) = definite integral from 0 to x : sin(t) / (t+1) dt.

    1) Find where f(x) has local maximum and minimum values (I should not find the function values).
    Max and min will occur where f'(x)= 0. And you find f' from the Fundamental Theorem of Calculus.

    A more interesting one:
    2) Show that f(x) > 0, when x > 0. (Here I should use integration by parts...)

    Do anyone want to help? :-)
    Jo, you should NOT use "integration by parts. You should not integrate at all. If f'(x)> 0 for x> 0, then f is increasing. and f(0)= ?
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  4. #4
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    Quote Originally Posted by Krizalid View Post
    since g(x)>0, g'(t)<0 and 1-\cos t<0, the whole RHS is positive and we're done.
    Hi, and thank you for helping me!
    But shouldn't it be 1-\cos t is equal or greater than 0, and not <0. ?? Then it makes sense to me...


    Quote Originally Posted by HallsofIvy View Post
    I have no idea! What was the question?
    Haha, I'm sorry! :-)
    This was originally a reply on another thread, but I think the moderator changed my answer to a new thread. I'm sorry.
    The question I was referring to is:
    http://www.mathhelpforum.com/math-he...em-162684.html

    And I got f(0) = 3. I was just wondering if anyone else got the same... :-)



    And again, thank you all for helping me!
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  5. #5
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    Quote Originally Posted by expresstrain View Post
    Hi, and thank you for helping me!
    But shouldn't it be 1-\cos t is equal or greater than 0, and not <0. ?? Then it makes sense to me...
    i got a typo, of course it is as you say.
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