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Homework Help: Evaluate integration

  1. Sep 3, 2004 #1
    Evaluate [tex]\int t^{14} ln(t) dt[/tex]


    using integration by parts and formula [tex]uv-\intv du[/tex]

    u=ln(t)
    du=1/t
    dv=t^14
    v=t^15/15

    lnt*t^15/15-[tex]\int[/tex]t^15/15*1/t

    lnt*t^15/15-t^14*log(t)

    i dont know what im doing wrong, please help
     
    Last edited: Sep 3, 2004
  2. jcsd
  3. Sep 3, 2004 #2

    jamesrc

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    First step is fine. You're over-thinking the second step:

    [tex] \int{\frac{t^{15}}{15} \frac{dt}{t}} = \int{\frac{t^{14}}{15}dt = \frac{t^{15}}{225} [/tex]

    So the answer should read:

    [tex] \int{\left(\ln t\right) t^{14} dt} = \frac{t^{15}}{15} \left(\ln t - \frac{1}{15} \right) [/tex]
     
    Last edited: Sep 3, 2004
  4. Sep 3, 2004 #3

    arildno

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    DO NOT USE THIS HORRIBLE MIXTURE OF INTEGRATION BY PARTS AND INTEGRATION BY SUBSTITUTION!!
    Write:
    [tex]\int{uv'}dt=uv-\int{u'v}dt[/tex]
    [tex]u=ln(t)\to{u}'=\frac{1}{t}[/tex]
    [tex]v'=t^{14}\to{v}=\frac{1}{15}t^{15}[/tex]
     
  5. Sep 3, 2004 #4

    what do you mean? your setup is just like mines
     
  6. Sep 5, 2004 #5

    arildno

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    Nope!
    1. "Using formula uv-du"
    Complete nonsense
    If you want to use this method properly, write:
    udv=d(uv)-vdu
    2.Furthermore
    you use a meaningless constuction like du=1/t
    If you want to use this method, you must write du=1/t*dt
    3. Your integral lacks dt, and is therefore meaningless as well
     
    Last edited: Sep 5, 2004
  7. Sep 6, 2004 #6

    ok i know how you got [tex] \int{\frac{t^{15}}{15} \frac{dt}{t}} = \int{\frac{t^{14}}{15}dt [/tex]

    but how did you get

    [tex]\int{\frac{t^{14}}{15}dt = \frac{t^{15}}{225} [/tex]

    i know you got the anti-derv. of it, and i tried to look in my book for a form like that, but i cant find it. can you explain to me how you did that?

    for [tex]\int{\frac{t^{14}}{15}dt[/tex] i used u-du sub.

    u=14
    du=u+1

    so... t^u/du but i cant seem to find that form in the book
     
  8. Sep 6, 2004 #7

    Dr Transport

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    [tex] \int \frac{t^{14}}{15} dt = \frac{1}{15}\int t^{14} dt = \frac{1}{15}\frac{t^{15}}{15} =

    \frac{t^{15}}{(15)^{2}} = \frac{t^{15}}{225} [/tex]
     
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