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P3 Integration Dude's Hard One! Who Can Crack It?! watch

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    question:

    integral of (lnx)^2dx

    with limits of e and 1
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    (Original post by mabs)
    question:

    integral of (lnx)^2dx

    with limits of e and 1
    is the answer 2 by any chance?!
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    e-2
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    (e - 2) ?
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    what method is everyone using - i don't know how to start
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    (Original post by mabs)
    question:

    integral of (lnx)^2dx

    with limits of e and 1
    2-ln1
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    (Original post by Silly Sally)
    what method is everyone using - i don't know how to start

    lol, intergrate by parts..... 2.lnx
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    (Original post by Silly Sally)
    what method is everyone using - i don't know how to start
    by parts
    u=(lnx)^2
    dv/dx = 1
    ...
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    u g2 use parts
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    (Original post by Bhaal85)
    2-ln1
    ln1 is 0, therefore answer is 2....which is wot i got, but its not right, lol
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    (Original post by Silly Sally)
    what method is everyone using - i don't know how to start
    lnx^2 can be done by parts, just do integration by parts with 1*lnx^2
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    IT STILL DOESN'T WORK!

    U GET 2/X lnx AND THAT WON'T REDUCE
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    (Original post by mabs)
    question:

    integral of (lnx)^2dx

    with limits of e and 1
    If you fell up to it you can integrate by parts

    u = lnx
    u' = 1/x
    v = xlnx + x
    v' = lnx

    integral of question = [ x(lnx)^2 - xlnx ] - [ integral of (lnx -1) ]

    => x(lnx)^2) - xlnx - xlnx +x +x

    then simplify and put the limits in and you get e-2. Maybe someone with html can type it out so it looks similar. Im sure theres an easier method but this one works!
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    check P3 ex 4d q.19
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    I got the right answer so here's the method:
    Integrate by parts with v = lnx, dv/dx = 1/x, du/dx = lnx, u = x(lnx - 1)
    So int (lnx)^2 = xlnx(lnx - 1) - int 1/x*(x(lnx - 1)) = x[(lnx)^2 - lnx] - x(lnx -1) + x = x[ln(x)^2 - 2 lnx + 2], evaluate between 1 and e to give (e - 2)
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    INT lnx^2 dx = INT 1*lnx^2

    u = lnx^2 therefore u' = 2/x
    v' = 1 therefore v = x

    xlnx^2 - INT x*2/x which gave me:

    [xlnx^2 - 2x] with limits e and 1

    [2e-2e] - [ln1-2]

    = 2-ln1
    = 2
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    George Explain!!!!!
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    BHAAL, (lnx)^2 means lnx all to the power 2 and u don't get wot u get when u use the chain rule.
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    (Original post by mabs)
    BHAAL, (lnx)^2 means lnx all to the power 2 and u don't get wot u get when u use the chain rule.
    It does work if you use the chain rule - thats the only bit i can do!!!
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    (Original post by Bezza)
    I got the right answer so here's the method:
    Integrate by parts with v = lnx, dv/dx = 1/x, du/dx = lnx, u = x(lnx - 1)
    So int (lnx)^2 = xlnx(lnx - 1) - int 1/x*(x(lnx - 1)) = x[(lnx)^2 - lnx] - x(lnx -1) + x = x[ln(x)^2 - 2 lnx + 2], evaluate between 1 and e to give (e - 2)

    i didnt know u could intergrate 'ln' functions?! how does du/dx + lnx become u = x(lnx - 1)???
 
 
 
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