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Does int_0^1 2x^3 ln(x)dx equal -1/8 or not

  1. Nov 4, 2005 #1
    I had an exam today and this was one of the problems:

    [tex] \int_{0}^{1} {2x^3 ln (x)} dx [/tex]

    My answer was -1/8. Is this correct or not?
     
  2. jcsd
  3. Nov 4, 2005 #2
    I got the integral to be...
    [tex] [(2x^{4}(lnx-1)-3x^{4}/2)/4] [/tex]
    ln 0 is infinate, thus the integral beteween 1 and 0 is infinate?!...
     
  4. Nov 4, 2005 #3
    Here is what I did (if I remember correctly):

    [tex] \int {2x^3ln(x)} dx [/tex]
    [tex]u = ln (x) [/tex] [tex] v' = 2x^3dx [/tex]
    [tex]u' = \frac {1} {x} [/tex] [tex] v = \frac {1} {2} x^4 [/tex]

    [tex]\int {2x^3ln(x)} dx = {\frac {1} {2} x^4} ln (x) - \int {{\frac {1} {x}}{\frac {1} {2} x^4}} dx = {\frac {1} {2} x^4} ln (x) - {\frac {1} {8} x^4} = x^4(\frac {1} {2} ln (x) - \frac {1} {8}) [/tex]
     
    Last edited: Nov 4, 2005
  5. Nov 4, 2005 #4
    My calculator got -1/8, and he is smart :smile: And your work looks correct too :smile:
     
  6. Nov 4, 2005 #5
    And then:

    [tex] \lim_{\substack{s\rightarrow 0^+}} ${\displaystyle x^4(\frac {1} {2} ln (x) - \frac {1} {8})|_{s}^{1}}$ = \lim_{\substack{s\rightarrow 0^+}} {[1^4(\frac {1} {2} ln (1) - \frac {1} {8})] - {[s^4(\frac {1} {2} ln (s) - \frac {1} {8})] = -\frac {1} {8} - 0 = -\frac {1} {8} [/tex]
     
    Last edited: Nov 4, 2005
  7. Nov 4, 2005 #6
    No need for all that complicated work

    You have x = 1, which means 1((1/2)ln(1) - 1/8) = -1/8
    You have x= 0, which means 0(something) = 0

    so the answer is -1/8
     
  8. Nov 4, 2005 #7
    Well, that's how they taught it. I'm sure I would be taken marks off if I didn't do it like that.

    BTW, what calculator do you have? I want one like that. :P
     
  9. Nov 4, 2005 #8
    Well, I don't know how your teacher grades, so you probably did the right thing by putting it in there. The Calculator is a TI-89.
     
  10. Nov 5, 2005 #9

    HallsofIvy

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    And they'd be perfectly correct to take marks off if you said:
    "You have x= 0, which means 0(something) = 0"

    0(something) is not always 0!
     
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