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Antiderivatives (where did I go wrong?)

  • Thread starter 01010011
  • Start date
  • #1
48
0
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. Homework 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?
 

Answers and Replies

  • #2
699
5
Cos of 0 equals 1 so 3 cos0=3
 
  • #3
699
5
And you subtracted 3*cos0 so it needs to be negative not positive.
 
  • #4
48
0
Cos of 0 equals 1 so 3 cos0=3
Thank you so much, I understand now!
 
  • #5
48
0
And you subtracted 3*cos0 so it needs to be negative not positive.
C = 5 - (0)^2 + 3 cos (0)
C = 5 - 0 - 3
C = 2

Ok, great , thanks
 
  • #6
699
5
Good except you still have +3 cos but the next steps are you have the negative sign
 
  • #7
48
0
Good except you still have +3 cos but the next steps are you have the negative sign
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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