Taylor's Theorem/One Variable/Integral Version

Proof
We first prove Taylor's theorem with the integral remainder term.

The Fundamental Theorem of Calculus states that


 * $\displaystyle \int_a^x f' \left({t}\right) \ \mathrm d t = f \left({x}\right) - f \left({a}\right)$

which can be rearranged to:


 * $\displaystyle f \left({x}\right) = f \left({a}\right) + \int_a^x f'(t) \ \mathrm d t$

Now we can see that an application of Integration by Parts yields:

Another application yields:
 * $\displaystyle f \left({x}\right) = f \left({a}\right)+(x-a) f' \left({a}\right)+ \frac 1 2 (x-a)^2f \left({a}\right) + \frac 1 2 \int_a^x (x-t)^2 f' \left({t}\right) \ \mathrm d t$

By repeating this process, we may derive Taylor's theorem for higher values of $n$.

This can be formalized by applying the technique of Principle of Mathematical Induction. So, suppose that Taylor's theorem holds for a $n$, that is, suppose that:

We can rewrite the integral using integration by parts. An antiderivative of $(x-t)^n$ as a function $t$ is given by $\dfrac{-(x-t)^{n+1}}{n+1}$, so:

The last integral can be solved immediately, which leads to


 * $\displaystyle R_n = \frac{f^{(n+1)}(\xi)}{(n+1)!} (x-a)^{n+1}$