Antiderivative problem! Given f'(x)=3cosx–11sinx and f(0)=–2,find f(x)=
f(x) = 3 sin x + 11 cos x + c
f(0) = -2 = 3 (0) + 11 (1) + c
so c = -13
f(x) = 3 sin x + 11 cos x - 13
To find the antiderivative of f'(x), we can use the method of integration. We know that the antiderivative of cosine is sine and the antiderivative of sine is negative cosine.
So, we integrate each term separately:
∫ 3cos(x) dx = 3∫ cos(x) dx = 3sin(x) + C1
∫ -11sin(x) dx = -11∫ sin(x) dx = -11(-cos(x)) + C2 = 11cos(x) + C2
Now, we add the two antiderivatives together:
f(x) = 3sin(x) + 11cos(x) + C
To find the value of C, we use the initial condition given in the problem:
f(0) = 3sin(0) + 11cos(0) + C = 0 + 11(1) + C = 11 + C = -2
Solving for C, we subtract 11 from both sides:
C = -2 - 11 = -13
Therefore, the final solution is:
f(x) = 3sin(x) + 11cos(x) - 13