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Definite integral (e^x) *(x-1)^n=16-6e find n.

  1. Sep 3, 2011 #1
    1. The problem statement, all variables and given/known data

    this is definite integral question.
    lower limit 0 ; upper limit 1 ; integral (e^x)(x-1)^n = 16-6e
    find n (n<6)
    here e is euler constant value around 2.7 (irrational)
    hope you understood.
    2. Relevant equations

    as much as first year student know.

    3. The attempt at a solution

    tried many substitutions like x-1 = t and some others.
    replaced x by 1-x and tried integration by parts. and many more.
    but failed:cry::yuck:
     
  2. jcsd
  3. Sep 3, 2011 #2
    Use integration by parts and you will be able to write the integral in summative notation
     
  4. Sep 3, 2011 #3
    does not understand.
    can you please explain it more briefly.
    thanks for reply.
     
  5. Sep 3, 2011 #4

    hunt_mat

    User Avatar
    Homework Helper

    Integrate by parts:
    [tex]
    \int_{0}^{1}e^{x}(x-1)^{n}dx=\Bigg[ e^{x}(x-1)^{n}\Bigg]_{0}^{1}-n\int_{0}^{1}e^{x}(x-1)^{n-1}dx=(-1)^{n}-n\int_{0}^{1}e^{x}(x-1)^{n-1}dx
    [/tex]
    If your original integral is [itex]I_{n}[/itex], then you have a recurrence relation:
    [tex]
    I_{n}=(-1)^{n}-nI_{n-1}
    [/tex]
     
  6. Sep 3, 2011 #5
    that's called answer thanks!!
    can you tell me how to use itex tex tags.
     
  7. Sep 3, 2011 #6

    hunt_mat

    User Avatar
    Homework Helper

    To open the tex just do tex in square brackets [ tex] (but without the spaces). To close of the tex do [ /tex ] but take off the spaces. The same with itex, which is in-line tex.
     
  8. Sep 3, 2011 #7
    As was pointed out:
    [tex]
    I_{n} \equiv \int_{0}^{1}{(x - 1)^{n} \, e^{x} \, dx}
    [/tex]

    [tex]
    I_{n} = \left. (x - 1)^{n} \, e^{x} \right|^{1}_{0} - n \int_{0}^{1}{(x - 1)^{n - 1} \, e^{x} \, dx}
    [/tex]

    [tex]
    I_{n} = (-1)^{n - 1} - n I_{n - 1}
    [/tex]

    [tex]
    I_{0} = \int_{0}^{1}{e^{x} \, dx} = e - 1
    [/tex]

    In general: [itex]I_{n} = A_{n} + B_{n} e[/itex]. The recursion gives:
    [tex]
    A_{n} = (-1)^{n - 1} - n A_{n - 1}
    [/tex]
    [tex]
    B_{n} = -n B_{n - 1}
    [/tex]
    with the initial conditions [itex]A_{0} = -1[/itex] and [itex]B_{0} = 1[/itex]. You need to find such an n that [itex]A_{n} = 16[/itex] and [itex]B_{n} = -6[/itex].
     
  9. Sep 3, 2011 #8
    The general solution to the B recursive relations is:
    [tex]
    B_{n} = (-1)^{n} n!
    [/tex]
     
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