Calculus-antiderivative

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Find f.
f ''(θ) = sin θ + cos θ, f(0) = 5, f '(0) = 3

My steps:
f'(θ)=cosθ-sinθ+C
When f'(0)=3, C=-2, so f'(θ)=cosθ-sinθ-2.

f(θ)=-sinθ-cosθ-2x+D
When f(0)=5, D=6, so f is -sinθ-cosθ-2x+6.

How is that wrong?

  • Calculus-antiderivative -

    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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