Answer:
f(x) = cos(x) - sin(x) + x/π
Step-by-step explanation:
First to know your antiderivatives, you must know your derivatives!
When you take a derivative of f(x) to get -sin(x), it must be cos(x), but if f'(x) is sin(x), f(x) must be -cos(x) to cancel out the negative. Likewise to get a derivative that is -cos(x), we must need a -sin(x). Thus in this problem we will take three steps:
- f'(x) = -sin(x) + -cos(x) + C1 (We end up with a constant of integration which occurs because the derivative of any constant is 0)
- f(x) = cos(x) + -sin(x) + C1 x + C2 (We tack on another constant of integration while combining our original with X because the second antiderivative contains an infinite amount of functions with the same derivative).
- Using our initial conditions f(0) = 1, and f(π)=0. We can substitute 0 into our function using our second constant at C1(0) which is always 0. This leads us to the equation: 1 = cos(0) - sin(0) + C2 -> 1 = 1 - 0 + C2 - > C2 = 0. Finally, we can find C1 through some substitution and algebra. 0 = cos(π) - sin(π) + C1x -> 0 = (-1) - (0) + C1(π) -> 1 = C1π -> C1 = 1/π.
Therefore, when we combine our information we get:
f(x) = cos(x) - sin(x) + x/π
We can also check that our solution works through substitution.
- f(0) = 1 = cos(0) - sin(0) + 0/π = 1 - 0 + 0 = 1 (Checks out).
- f(π) = 0 = cos(π) - sin(π) + π/π = -1 - 0 + 1 = 0 (Checks out).
