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Integration Problem

  1. Feb 14, 2008 #1
    1. The problem statement, all variables and given/known data

    I don't know how to make the integral sign :frown: so I am took a picture of the question.

    [​IMG]


    2. Relevant equations

    Equations are located in the above img.


    3. The attempt at a solution

    I first integrated the first derivative and got -

    (ax^3)/3 + (bx^2)/2

    plugged in 1 and got -

    6=a + b

    Then I got the second derivative --> 2ax +b and plugged in f''(1) and got: 18 = 2a +b

    Using both the equations, I figured out that:
    a=12
    b=-6

    how do I find the contant c to get f(x)?


    -Jennie
     
  2. jcsd
  3. Feb 14, 2008 #2

    EnumaElish

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    By using (iii)?

    I am assuming you meant f(x) = (ax^3)/3 + (bx^2)/2 + c.
     
  4. Feb 14, 2008 #3
    I tried using (iii) and this is what I got -

    f(x) = 4x^3 -3x^2 + c

    then I plugged in 2 and 1 and got -

    (32-12+c) - (4-3+c) = 18

    but then c will cancel out.

    I am sure I am doing this wrong. please guide me
     
  5. Feb 14, 2008 #4

    EnumaElish

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    What do you get when you integrate 4x^3 -3x^2 + c without limits?
     
  6. Feb 14, 2008 #5
    x^4 - x^3 + cx
     
  7. Feb 14, 2008 #6

    EnumaElish

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    Shouldn't it be x^4-x^3+cx?
     
  8. Feb 14, 2008 #7
    yes, i edited it before you replied.
     
  9. Feb 14, 2008 #8

    EnumaElish

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    What do you get for x=1? For x = 2?
     
  10. Feb 14, 2008 #9
    when x=1 then I got c
    when x=2 then I got 8+2c
     
  11. Feb 14, 2008 #10

    EnumaElish

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    Do you think you can use (iii)?
     
  12. Feb 14, 2008 #11
    Thank you!

    so,

    (8 +2c) - c = 18
    and c = 10

    that gives me--> f(x)= 12x^2 - 6x + 10
     
  13. Feb 14, 2008 #12

    EnumaElish

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    Do you mean f(x) = 4x^3 -3x^2 + 10?
     
  14. Feb 14, 2008 #13
    aaah yes. I was writing it down and was wondering how 6=18.

    Sorry for the error and Thanks for the help. :)
     
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