Given f'(x)=3cosx-10sinx and f(0)= -6
Find f(x)
To find f(x), we need to integrate f'(x). Since f'(x) = 3cos(x) - 10sin(x), we will integrate each term separately.
∫[f'(x) dx] = ∫[(3cos(x) - 10sin(x)) dx]
To integrate the first term, ∫[3cos(x) dx], we can use the fact that the integral of cos(x) is sin(x) plus a constant:
∫[3cos(x) dx] = 3∫[cos(x) dx] = 3sin(x) + C1
To integrate the second term, ∫[-10sin(x) dx], we can use the fact that the integral of sin(x) is -cos(x) plus a constant:
∫[-10sin(x) dx] = -10∫[sin(x) dx] = -10(-cos(x)) + C2 = 10cos(x) + C2
Therefore, the integral of f'(x) is:
∫[f'(x) dx] = 3sin(x) + C1 + 10cos(x) + C2
Since we are given f(0) = -6, we can substitute this value to find the values of the constants C1 and C2:
f(0) = 3sin(0) + C1 + 10cos(0) + C2 = C1 + C2
Since f(0) = -6, we have:
C1 + C2 = -6
Substituting the values of C1 and C2 back into the integral expression, we have:
f(x) = 3sin(x) + 10cos(x) + (C1 + C2)
f(x) = 3sin(x) + 10cos(x) - 6
To find f(x), we will integrate f'(x) with respect to x.
∫[f'(x)]dx = ∫[3cos(x) - 10sin(x)]dx
To integrate cos(x), we use the identity: ∫cos(x)dx = sin(x) + C1, where C1 is the constant of integration.
To integrate sin(x), we use the identity: ∫sin(x)dx = -cos(x) + C2, where C2 is another constant of integration.
So, integrating f'(x) gives us:
f(x) = 3∫cos(x)dx - 10∫sin(x)dx
= 3(sin(x) + C1) - 10(-cos(x) + C2)
Now, we need to find the values of C1 and C2. To do that, we will use the initial condition given: f(0) = -6.
Substituting x = 0 into the equation f(x):
-6 = 3(sin(0) + C1) - 10(-cos(0) + C2)
-6 = 3(0 + C1) - 10(-1 + C2)
-6 = 3C1 + 10C2 - 10
Simplifying the equation above, we get:
3C1 + 10C2 = 4 ......(1)
Now, we can substitute the values of C1 and C2 back into the equation for f(x):
f(x) = 3sin(x) + 3C1 + 10cos(x) - 10C2
And that is the final expression for f(x).