Antiderivative question (pls help)

Question

Find f.



f '''(x) = cos(x),    f(0) = 3,    f '(0) = 8,    f ''(0) = 4

Answer 1

f"(x) = ∫f'''(x)dx = ∫cosxdx = sinx + C₁

Since f"(0) = 4, we have sin0 + C₁ = 4. So, C₁ = 4

f"(x) = sinx + 4

f'(x) = ∫f"(x)dx = ∫(sinx + 4)dx = -cosx + 4x + C₂

Since f'(0) = 8, -cos0 + 4(0) + C₂ = 8. So, C₂ = 9

f'(x) = -cosx + 4x + 9

f(x) = ∫f'(x)dx = ∫(-cosx + 4x + 9)dx = -sinx + 2x² + 9x + C₃

Since f(0) = 3, -sin0 + 2(0)² + 9(0) + C₃ = 3. So, C₃ = 3

Therefore, f(x) = -sinx + 2x² + 9x + 3

Answer 2

Just have to do one at a time.

f'''(x) = cos(x)

f''(x) = sin(x) +C

f''(0) = 4

sin(0) + C = 4

C = 4

f''(x) = sin(x) +4

f'(x) = -cos(x) + 4x + C

f'(0) = 8

-cos(0) + 4(0) + C = 8

-1 + C = 8

C = 9

f'(x) = -cos(x) + 4x + 9

f(x) = -sin(x) + 2x² + 9x + C

^{f(0) = 3}

-sin(0) +2(0)² + 9(0) + C = 3

C = 3

f(x) = -sin(x) + 2x² + 9x + 3