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Integration of a polynomial problem

  1. Nov 7, 2014 #1
    Hi,
    I'm using KA Stroud 6th edition (for anyone with the same book, P407) and there is a example question where I just can't seem to get the answer they have suggested:

    1. The problem statement, all variables and given/known data

    Question:
    Determine the value of I = ∫(4x3-6x2-16x+4) dx
    when x = -2, given that at x = 3, I = -13
    Their answer is when x = -2, I = 12.

    3. The attempt at a solution

    I found the integral: I = ∫(4x3-6x2-16x+4) dx = x4-2x3-8x2+4x + C
    and then substituted x for 3 and getting:
    -13 = -33 + C
    thus:
    C = 20
    Now when I replace x with -2, plus the constant, I get:
    -24-2(-2)3-8(-2)2+4(-2) + 20 = -20

    I'm a few days into Integrals so I feel I may be doing something daft?

    Many thanks.
     
  2. jcsd
  3. Nov 7, 2014 #2

    Mark44

    Staff: Mentor

    Your work is fine except for one minor thing. At the end you wrote -24 instead of (-2)4. In the first, 2 is raised to the 4th power, and then you take the negative, resulting in -16. In the latter, -2 is raised to the 4th power, resulting in +16.
     
  4. Nov 7, 2014 #3
    That was an honest slip. I appreciate there is a difference between them both.
    I finally got the answer, it was a case of not respecting the brackets and powers... I need a break.

    Thanks for pointing that out and explaining! :)
     
  5. Nov 7, 2014 #4

    LCKurtz

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    HEY!! I thought micromass's avatar was retired.
     
  6. Nov 8, 2014 #5
    I assume that is for me? :). I use this logo for most online things. EEVBlog, my Steam account etc etc. I was thinking about using Floyds new album cover.
     
  7. Nov 8, 2014 #6

    LCKurtz

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    Yes, but it was really directed at the old timers, sort of tongue-in-cheek. Turns out one of our previous highly regarded members used to use that logo. Nothing to worry about though.
     
  8. Nov 9, 2014 #7
    Yeah, that's cool. I Understand :).
     
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