Calculusantiderivative
posted by Callie .
Find f.
f ''(θ) = sin θ + cos θ, f(0) = 5, f '(0) = 3
My steps:
f'(θ)=cosθsinθ+C
When f'(0)=3, C=2, so f'(θ)=cosθsinθ2.
f(θ)=sinθcosθ2x+D
When f(0)=5, D=6, so f is sinθcosθ2x+6.
How is that wrong?

Calculusantiderivative 
Reiny
f'Ø) = cosx + sinx + C
you differentiated instead of integrated.
Had you checked your answer by taking the derivative you would have seen your error
since f'(0) = 3
3 = cos0 + sin0 + c
3 = 1 + 0 + c > c = 4
so f'(x) = cosx + sinx + 4
f(x) = sinx  cosx + 4x + d
and f(0) = 5
5 = sin0  cos0 + 0 + d
5 = 0  1 + d
d = 6
f(x) = sinx  cosx + 4x + 6
Just realized I used x instead of Ø, no big deal to change
I suggest you take a better look at the derivatives of both sine and cosine
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