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Math Help - just need explanation for integration prob.

  1. #1
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    just need explanation for integration prob.

    Ok so this is the problem:

    Integral from 1 to e of (ln(x))/x) dx (sorry don't know how to put the intergral symbol in)

    I set u= lnx and du= 1/x and changed the limits to 0 to 1
    When i solve though, I get 1, and my book says the answer is 1/2. I've tried this a bunch of times but can't get their answer. Any explanation?
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  2. #2
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    Quote Originally Posted by painterchica16 View Post
    Ok so this is the problem:

    Integral from 1 to e of (ln(x))/x) dx (sorry don't know how to put the intergral symbol in)

    I set u= lnx and du= 1/x and changed the limits to 0 to 1
    When i solve though, I get 1, and my book says the answer is 1/2. I've tried this a bunch of times but can't get their answer. Any explanation?
    So after your sub you get

    \int_{0}^{1}u du =\frac{1}{2}u^2 \bigg|_{0}^{1}=\frac{1}{2}(1^2-0^2)=\frac{1}{2}
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  3. #3
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    \int_{1}^{e} \frac{lnx}{x}

    u = lnx

    du = 1/x

    \int udu

    Raise power and divide by power.

    \int udu = \frac{1}{2}u^2

    \frac{(lnx)^2}{2}

    From 1 to e,

    1/2ln(e)^2 - 1/2ln(1)^2

    ln(1) = 0, ln(e) = 1
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  4. #4
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    Thanks, I understand the setup now, but the only that I'm still confused about is why you plugged 0 and 1 into just u^2, and not lnx?

    Because, ln(1)^2=0 and ln(0)^2 just doesnt exist. Why does it not work when you plug it in?
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  5. #5
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    If u = lnx, than our intervals are no longer 1 and e but ln(1) and ln(e). Hence the substitution.
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  6. #6
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    Oh, I get it, you plug it into u^2, not the other thing, because its all in terms of u. Thanks!
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