# Explain why the Taylor Remainder Theorem in Integral Form holds true.

For instance:

Let $f(x)=\sin x$ and $a=\pi/2$. Explain why the Taylor Remainder Theorem in Integral form holds; that is, why we can write $f(\pi/2+1/2)-P_4(\pi/2+1/2)=E_4(\pi/2+1/2)$ where $E_4(\pi/2+1/2)$ is the error in integral form.

I'm having a bit of trouble starting the proof. I know that $E_1(x)=\int_0^x(x-t)f''(t)dt$ and similarly, $f(x)=\sum_{k=0}^n{f^{(k)}(a)\over{k!}}(x-a)^k+E_n(k)$ where $E_n(x)={1\over{n!}}\int_a^x(x-t)^nf^{n+1}(t)dt$. However, I'm not sure where to go from here except by maybe using $n=4$ and proving the above. It seems a bit reductive and self-evident. I'm wondering how anyone else would do this proof. Thanks.

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I seem to recall that you integrate by parts: $$f (x)-f (a)=\int_a^x f '(t)\,dt.$$ Then \begin{align} f (x)-f (a)&=\int_1^x f '(t)\,dt=-\left.\vphantom {\int}(x-t)f'(t)\right|_a^x+\int_a^x (x-t)f''(t)\,dt\\ \ \\ &=(x-a)f'(a)+\int_a^x (x-t)f''(t)\,dt. \end{align}