Editing Taylor's theorem (section)

=== Derivation for the integral form of the remainder ===

Due to the [[absolutely continuous|absolute continuity]] of <math>f^{(k)}</math> on the [[closed interval]] between <math display=inline>a</math> and <math display=inline>x</math>, its derivative <math>f^{(k+1)}</math> exists as an <math>L^1</math>-function, and we can use the [[fundamental theorem of calculus]] and [[integration by parts]]. This same proof applies for the [[Riemann integral]] assuming that <math>f^{(k)}</math> is [[continuous function|continuous]] on the closed interval and [[Differentiable function|differentiable]] on the [[open interval]] between <math display=inline>a</math> and <math display=inline>x</math>, and this leads to the same result as using the mean value theorem.

The [[fundamental theorem of calculus]] states that

<math display="block"> f(x)=f(a)+ \int_a^x \, f'(t) \, dt.</math>

Now we can [[Integration by parts|integrate by parts]] and use the fundamental theorem of calculus again to see that

<math display="block"> \begin{align}
f(x) &= f(a)+\Big(xf'(x)-af'(a)\Big)-\int_a^x tf''(t) \, dt \\
&= f(a) + x\left(f'(a) + \int_a^x f''(t) \,dt \right) -af'(a)-\int_a^x  tf''(t) \, dt \\
&= f(a)+(x-a)f'(a)+\int_a^x \, (x-t)f''(t) \, dt,
\end{align} </math>

which is exactly Taylor's theorem with remainder in the integral form in the case <math>k=1</math>. The general statement is proved using [[mathematical induction|induction]]. Suppose that
{{NumBlk|:|<math> f(x) = f(a) + \frac{f'(a)}{1!}(x - a) + \cdots + \frac{f^{(k)}(a)}{k!}(x - a)^k + \int_a^x \frac{f^{(k+1)} (t)}{k!} (x - t)^k \, dt. </math>|{{EquationRef|eq1}}}}

Integrating the remainder term by parts we arrive at

<math display="block">\begin{align}
\int_a^x \frac{f^{(k+1)} (t)}{k!} (x - t)^k \, dt = & - \left[ \frac{f^{(k+1)} (t)}{(k+1)k!} (x - t)^{k+1} \right]_a^x + \int_a^x \frac{f^{(k+2)} (t)}{(k+1)k!} (x - t)^{k+1} \, dt \\
= & \ \frac{f^{(k+1)} (a)}{(k+1)!} (x - a)^{k+1} + \int_a^x \frac{f^{(k+2)} (t)}{(k+1)!} (x - t)^{k+1} \, dt.
\end{align}</math>

Substituting this into the formula {{nowrap|in ({{EquationNote|eq1}})}} shows that if it holds for the value <math>k</math>, it must also hold for the value <math>k+1</math>. Therefore, since it holds for <math>k=1</math>, it must hold for every positive integer <math>k</math>.