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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?
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?