Final answer:
The integral ∫ [f(x) + f''(x)]sin(x)dx can be evaluated using integration by parts, considering the conditions f(a) = f'(a) = 0 provided in the question. Without specific functions for f(x), we reach an expression in terms of the antiderivatives of f'(x) and f'''(x) times cos(x), which cannot be further simplified without additional boundary conditions.
Step-by-step explanation:
To evaluate the integral ∫ [f(x) + f''(x)]sin(x)dx, we can use integration by parts, which in its general form is ∫ u dv = uv - ∫ v du. In this case, let u = f(x) + f''(x) and dv = sin(x)dx. This choice is convenient because the derivatives of sine and cosine functions are cyclic. However, we're given that f(a) = f'(a) = 0, which simplifies our calculations, as the boundary terms will vanish when we apply these conditions.
Let's perform integration by parts:
- u = f(x) + f''(x)
- dv = sin(x)dx
- du = (f'(x) + f'''(x))dx
- v = -cos(x)
Applying integration by parts, we have:
-∫ [f(x) + f''(x)]cos(x)dx - (f(x) + f''(x))cos(x) | evaluated from 0 to a
Given f(a) = f'(a) = 0, the boundary terms at 'a' are zero. If we also assume that a similar condition holds at zero, which seems to be suggested by the question, the boundary terms at 0 would also be zero. Hence the value of the integral would be:
- ∫ [f'(x) + f'''(x)]cos(x)dx. The integral of the sum can be split into the sum of integrals:
-∫ f'(x)cos(x)dx - ∫ f'''(x)cos(x)dx.
Without specific functions provided for f(x), we cannot evaluate the integral further. However, if additional boundary conditions are known for f'(0) or f''(0), further simplifications could be made. As no boundaries other than f(a) and f'(a) equal zero were given, we can only say the integral is equal to an expression in terms of the antiderivatives of f'(x)cos(x) and f'''(x)cos(x) evaluated from 0 to a.