Consider $f(x) = f(0) + \int_0^xf'(t)dt$. I must show, through integrating $f(0) + \int_0^xf'(t)dt$ by parts, that $f(x) = f(0) + f'(0)x + \int_0^x(x-t)f''(t)dt$. We are also told that f is an infnitely differentiable function.
My attempt:
we must show $\int_0^xf'(t)dt = f'(0)x + \int_0^x(x-t)f''(t)dt$. Using integration by parts, we get that $\int_0^xf'(t)dt = f'(t)t + \int_0^x-tf''(t)dt$.
This is about as far as I get. I've tried some other things but they don't really seem to work.
You dropped your integration bounds, so from your integration by parts, you should actually get $$ \int_0^x{f'(t) dt} = f'(t)t\bigr|_0^x - \int_0^x{tf''(t)dt}= f'(x)x-\int_0^x{tf''(t)dt}$$ The next thing to notice is that by the Fundamental Theorem of Calculus, $$f'(x)x-f'(0)x = \int_0^x{xf''(t)dt}$$ Thus, we find $$\int_0^x{f'(t)dt} = f'(0)x +\int_0^x{xf''(t)dt} - \int_0^x{tf''(t)dt}= f'(0)x+\int_0^x{(x-t)f''(t)dt}$$