This image of part of a proof for the Lagrange Remainder for Taylor's Formula. I need help solving the last integral. Can anyone explain?
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The last integral can't really be "solved immediately." If it could be, then so could the integral they rewrote with integration by parts, as they're the exact same integral.
Here is an alternate, cleaner, and clear way to finish their proof.
We have the integral $$\int_a^x \frac{f^{(n+1)}(t)}{n!}(x-t)^n \mathrm dt.$$
By the mean value theorem for integrals (which is really the intermediate value theorem in disguise), we can rewrite this as
$$ f^{(n+1)}(\xi) \int_a^x \frac{(x-t)^n}{n!} \mathrm dt = f^{(n+1)}(\xi) \frac{(x-a)^{n+1}}{(n+1)!},$$
where $\xi$ is some number between $a$ and $x$, and where the last equality comes from evaluating the (now actually easy to evaluate) integral. $\diamondsuit$
answered Aug 20, 2014 at 6:03
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$\begingroup$ any idea why the proof in the OP would bother with the I.B.P. before saying the resulting integral could be “solved immediately”? I don’t see any advantage to delaying the step you provided by their penultimate integration. Thanks for pointing to the Wiki article. With your permission, I’d add an edit with the relevant summary of the MVT, as it applies here, to complete the answer as a stand-alone reference. Very much appreciated! $\endgroup$ Jun 25, 2019 at 16:43
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$\begingroup$ @RaxAdaam I don't know why the OP's proof uses IBP. It does not seem helpful. And feel free to improve my answer through any edits you deem appropriate. $\endgroup$– davidlowryduda ♦Jun 25, 2019 at 19:40