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Math Help - integral

  1. #1
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    integral

    Hi everyone,

    Could someone tell me if this is correct so far, please?

    Integral 0 to 1 lnx/sq. x

    lim t goes to 1 from the left lnx/sq. x

    u=lnx
    dv=sq. x
    du=1/x
    v=1/2xto the -1/2

    lim t goes to 1 from the left lnx(1/2x^-1/2)-int.(1/2x^-1/2)(1/x)

    Could I break this up into two integrals?
    -1/2int. x^-1/2-1/2int. 1/x

    Thank you very much
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  2. #2
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    Quote Originally Posted by chocolatelover View Post
    Hi everyone,

    Could someone tell me if this is correct so far, please?

    Integral 0 to 1 lnx/sq. x

    lim t goes to 1 from the left lnx/sq. x

    u=lnx
    dv=sq. x
    du=1/x
    v=1/2xto the -1/2

    lim t goes to 1 from the left lnx(1/2x^-1/2)-int.(1/2x^-1/2)(1/x)

    Could I break this up into two integrals?
    -1/2int. x^-1/2-1/2int. 1/x

    Thank you very much
    the function is undefined at 0, so you would find \lim_{t \to 0} \int_{t}^{1} \frac {\ln x}{\sqrt {x}}~dx

    Let u = \ln x and dv = \frac 1{\sqrt {x}} = x^{-1/2}
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  3. #3
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    Does -1/2 look right?

    At the end I got the lim as t goes to 1-[(-1^1/2)-[ln(0)(0^1/2)/2)-0^1/2]

    So, that would just be -1/2, right?

    Thank you very much
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by chocolatelover View Post
    Does -1/2 look right?

    At the end I got the lim as t goes to 1-[(-1^1/2)-[ln(0)(0^1/2)/2)-0^1/2]

    So, that would just be -1/2, right?

    Thank you very much
    no. we spoke about this already. where is your t? and i told you that the limit is as t goes to zero. why is it you have trouble following clear instructions?

    you should have: \lim_{t \to 0} \left[ 2 \sqrt {x} \ln x - 4 \sqrt {x} \right]_{t}^{1}
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  5. #5
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    oops!
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  6. #6
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    Is it 0 from the left or 0 from the right?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by chocolatelover View Post
    Is it 0 from the left or 0 from the right?
    from the right, this is what the graph looks like

    this is my 41th post!!!
    Attached Thumbnails Attached Thumbnails integral-graph31.jpg  
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  8. #8
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    I got -4. Is that correct?

    Thank you
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  9. #9
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    Quote Originally Posted by chocolatelover View Post
    I got -4. Is that correct?

    Thank you
    yup, that's what i got. good job
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