- #1

01010011

- 48

- 0

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

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

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

= [(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?