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2 C4 integration questions im stuck on watch

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    Hi all,
    Can anyone help me out with these two questions, I spent some time on them then gave up and moved on but now their really starting to bug me so I would appreciate some help. Thanks

    Use substitution u=lnx to show that Integral(e^2x to e) 1/x(sqrt)lnx dx. is equal to 2 root 2 -1

    Sorry hope u can understand what ive written, im trying to learn to use latex
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    (Original post by Virgil)
    Hi all,
    Can anyone help me out with these two questions, I spent some time on them then gave up and moved on but now their really starting to bug me so I would appreciate some help. Thanks

    Use substitution u=lnx to show that Integral(e^2x to e) 1/x(sqrt)lnx dx. is equal to 2 root 2 -1

    Sorry hope u can understand what ive written, im trying to learn to use latex
    Sorry, I can't follow that. If your Latex isn't up to it, try specifying the limits and what you're integrating separately, and liberal use of brackets to make things clear.
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    Okay well the limits of the integral are e^2x to e
    The bit after the integral is (1) divided by the ((square root of (lnx) times) times x) dx.

    And the answer is suppose to be (2root2)-2

    Hope thats abit clearer thank you
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    I'm somewhat puzzled by your lower limit: e^2x is a function of x, or am I being particularly thick at the moment.
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    that what is says on the question Integral from e^2x at the top to e at the bottom
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    idiots
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    OK, this is what I think the question should be:

    \displaystyle\int^{e^2}_{e}\frac  {1}{x\sqrt{\ln x}}\;\;\text{dx}

    and the answer is 2(\sqrt{2}-1)

    Does that seem reasonable from what you've got?

    If so, how far have you got?
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    So first ive found that u=lnx so du/dx=1/x therefore dx=x.du so now I replace the dx with x.du

    So my equation so far is the integral of limits 1/x(sqrt u) times by x.du

    So the x cancels on top and bottom to leave 1/(sqrt U) du. Rights so far?

    This is the bit im stuck on how do I change the limits so that they are in terms of du and not dx?
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    and the answer should have the (-1) outside the bracket of the 2root2 bit
    The equation bit is right
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    sorry mu fault i miss read the question YOU are RIGHT it cant be e^2x it is e^2 SORRY
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    (Original post by Virgil)
    So the x cancels on top and bottom to leave 1/(sqrt U) du. Rights so far?
    That's the correct integral.

    This is the bit im stuck on how do I change the limits so that they are in terms of du and not dx?
    I'll rephrase that "in terms of u and not x".

    Just use the substitution "u= ln x". E.g. for the upper limit you'll have \ln(e^2), which is...?
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    ah right so now the upper limit is just 2 and the lower limit is now 1
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    (Original post by ghostwalker)
    That's the correct integral.



    I'll rephrase that "in terms of u and not x".

    Just use the substitution "u= ln x". E.g. for the upper limit you'll have \ln(e^2), which is...?
    Im a little confused as to how to go about the next step, the integral of 1/(root)u .
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    U dont use the ln rule and ive tried just integrating normally and I dont get 2root2 -1
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    (Original post by Virgil)
    Im a little confused as to how to go about the next step, the integral of 1/(root)u .
    You'll kick yourself!

    Rewritiing it as \int u^{-\frac{1}{2}}\;\;\text{du}

    Can you see?
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    (Original post by Virgil)
    U dont use the ln rule
    I don't understand what you are talking about there.

    Also I never said you got to 2root2 -1. I got to 2(root2-1), as per my previous posting.

    Edit: I've just run it through some graphing software, and if that integral is correct, then the answer is 2(root2-1), and not 2root2-1
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    FML the answer in the book says it is 2root2 -2 not -1. FML
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    (Original post by Virgil)
    U dont use the ln rule and ive tried just integrating normally and I dont get 2root2 -1
    I follow you now. Yes, you want to integrate normally, (ln rule not required). The answer is 2root2 -2 or 2(root2-1)
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    (Original post by ghostwalker)
    I don't understand what you are talking about there.

    Also I never said you got to 2root2 -1. I got to 2(root2-1), as per my previous posting.

    Edit: I've just run it through some graphing software, and if that integral is correct, then the answer is 2(root2-1), and not 2root2-1
    The ln rule is where f(x) divided by f'(x) = ln(f(x)) +c

    Sorry buddy
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    Cool.
 
 
 
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