# Antiderivatives (where did I go wrong?)

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

Cos of 0 equals 1 so 3 cos0=3

And you subtracted 3*cos0 so it needs to be negative not positive.

Cos of 0 equals 1 so 3 cos0=3

Thank you so much, I understand now!

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

Good except you still have +3 cos but the next steps are you have the negative sign

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