Taylor expansion, integration by parts, and the integration of dt. So my notes say, for a continuous function we have
$$ \int_a^x f'(t)dt = f(x) - f(a) \tag 1 $$
which I understand.  So re-arranging gives.
$$ f(x) = f(a) + \int_a^x f'(t)dt \tag 2 $$
or
$$ f(x) =f(a) + R_o(x) \tag 3$$
Which I believe is like saying $ f(x) $ is equal to $ f(a) $ plus some remainder.  My notes now say, we can integrate $R_o(x)$ by parts, and states,
$$ u= f'(t), du=f''(t) \tag 4$$
$$ dv = dt, v=-(x-t) \tag 5$$
Firstly, I am confused as to the integration by parts.  Why is there a need for this?  As far as I can see, there is only one part, $f'(t)$.
Secondly, if I am to treat it as two parts, $f'(t)$ and $dt$, then why does $v= -(x-t)$ and not simply $t$.
Thank you.
 A: To see why you need to set $v = -(x-t)$ you need to actually do the integration by parts.
Suppose you let $v=t$, then when you performed the integration by parts, you would end up with the useless result$$f(x) = f(a) +xf'(x) - af'(a) -\int_a^xf''(t)tdt$$ By setting $v = -(x - t)$ we get the following...
$$f(x) = f(a) + \int_a^xf'(t)dt$$
$$f(x) = f(a) + f'(x)[-(x-x)] - f'(a)[-(x-a)] - \int_a^xf''(t)[-(x-t)]dt$$
$$f(x) = f(a) + (x-a)f'(a) + \int_a^x(x-t)f''(t)dt$$
Now we are to the point where we can actually see two parts ($f''(t)$ and $(x-t)$). Integrating by parts again (with $dv = ((x-t)/1!)dt$ and $v = -(x-t)^2/2!$) we have
$$f(x) = f(a) + (x-a)f'(a) + \frac{(x-a)^2}{2!}f''(a) + \int_a^x\frac{(x-t)^2}{2!}f'''(t)dt$$
If we continue this process (successively setting $v = -(x-t)^2/2!$, then $-(x-t)^3/3!$, etc.) we end up with the following
$$f(x) = \sum_{k=0}^n\frac{(x-a)^k}{k!}f^{(k)}(a) + \int_a^x\frac{(x-t)^n}{n!}f^{(n+1)}(t)dt$$
Don't just read your notes passively, be active ... it can deepen your understanding :)!
