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Math Help - two intregal problems

  1. #1
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    two intregal problems

    Hi all,

    can someone solve these two problems for me?

    The first one is to proof is that is true or not

    and the second one, I just totally forgot how to do the anti derivative of that....

    two intregal problems-math.jpg

    Thanks!!!

    [IMG]file:///C:/DOCUME%7E1/UR_PUB%7E1.UR/LOCALS%7E1/Temp/msohtmlclip1/01/clip_image002.gif[/IMG]
    [IMG]file:///C:/DOCUME%7E1/UR_PUB%7E1.UR/LOCALS%7E1/Temp/msohtmlclip1/01/clip_image002.gif[/IMG]



    [IMG]file:///C:/DOCUME%7E1/UR_PUB%7E1.UR/LOCALS%7E1/Temp/msohtmlclip1/01/clip_image002.gif[/IMG]
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  2. #2
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    Here's a hint

    \int \frac{1}{t^2+1}~dt = \arctan (t) +C
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  3. #3
    Member Jskid's Avatar
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    With respect to the first question: I think the first step is to integrate both functions
    for \int_0^1 \sqrt{1+x^2} sinces it's of the form \sqrt{a^2+x^2} let x=\arctan\theta
    and for int_0^1 \sqrt{1+x} use u-substitution with u=1+x

    I'm not sure but integration may note be nescecary...because the one with x^2 is obviously always going to be greater than or equal to the other.

    Does this help?
    Last edited by Jskid; July 14th 2010 at 05:46 PM. Reason: does my explination as to why you may not need to integrate make sense?
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  4. #4
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    Quote Originally Posted by Jskid View Post
    I'm not sure but integration may note be nescecary...because the one with x^2 is obviously always going to be greater than or equal to the other.
    But after you integrate that will change.
    Last edited by TheCoffeeMachine; July 14th 2010 at 07:01 PM.
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  5. #5
    Member mohammadfawaz's Avatar
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    Quote Originally Posted by J
    I'm not sure but integration may note be nescecary...because the one with [tex
    x^2[/tex] obviously always going to be greater than or equal to the other.

    Does this help?
    yeah...clearly integration is not needed. Simply notice that x is between 0 and 1 and hence x^2\leq x. This directly leads to the given inequality since everything else is the same in both sides.
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  6. #6
    A Plied Mathematician
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    jskid and TheCoffeeMachine are both incorrect on some points; mohammadfawaz is correct. On the interval from 0 to 1, x^{2}\le x. That's true regardless of whether you integrate or not.
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  7. #7
    MHF Contributor chisigma's Avatar
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    The first question doesn't require any integration tool because for 0 \le x \le 1 is \sqrt{1+x^{2}} \le \sqrt{1+x} ...

    Kind regards

    \chi \sigma
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