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Homework Help: Antiderivatives (where did I go wrong?)

  1. Mar 20, 2010 #1
    Where did I go wrong with my working? The answer in the book is
    f(t) = t^2 + 3 cos t + 2

    1. The problem statement: Find f for
    f prime (t) = 2t - 3 sin t, f(0) = 5


    2. Relevant equations:
    Most General antiderivation: F(x) + C
    Antidifferentiation formula: Function = sin x, Particular antiderivation = -cos x


    3. My attempt at a solution

    = [(2t^2)/2] -3 (-cos) t + C

    = t^2 + 3 cos t + C

    f(0) = t^2 + 3 cos t + C = 5

    Therefore C = 5 - (0)^2 + 3 cos (0)
    C = 5

    Therefore f(t) = t^2 - 3 cos t + 5


    Why is this answer wrong?
     
  2. jcsd
  3. Mar 20, 2010 #2
    Cos of 0 equals 1 so 3 cos0=3
     
  4. Mar 20, 2010 #3
    And you subtracted 3*cos0 so it needs to be negative not positive.
     
  5. Mar 20, 2010 #4
    Thank you so much, I understand now!
     
  6. Mar 20, 2010 #5
    C = 5 - (0)^2 + 3 cos (0)
    C = 5 - 0 - 3
    C = 2

    Ok, great , thanks
     
  7. Mar 20, 2010 #6
    Good except you still have +3 cos but the next steps are you have the negative sign
     
  8. Mar 20, 2010 #7
    Careless mistakes on my part.

    C = 5 - (0)^2 - 3 cos (0)
    C = 5 -(3 * 1)
    C = 5 - 3
    C = 2
     
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