Remainder in taylor formula

Question

I found on a book a version of Taylor's formula like this:
$f\big(X(t)\big)-f\big(X(s)\big)=f^\prime\big(X(s)\big)\big(X(t)-X(s)\big)+\frac{1}{2}f^{\prime\prime}\big(X(s)\big)\big(X(t)-X(s)\big)^2+\int_0^1(1-\alpha)\Big[f^{\prime\prime}\big((1-\alpha)X(t)+\alpha X(s)\big)-f^{\prime\prime}\big(X(s)\big)\Big]\big(X(t)-X(s)\big)^2 d\alpha$
How can one prove such a formula? I expected to have a derivative of order 3 in the remainder. (actually the dependance of $X$ on $t$ or $s$ here is of no importance, my question would be the same with $X(t)=x$ and $X(s)=y$)
Thanks

Answer 1

I now recognize it is quite a silly question: you just write taylor expansion to the first order with integral remainder, change variables in the integral and then add and subtract
$\frac{1}{2}f^{\prime\prime}\big(X(s)\big)\big(X(t)-X(s)\big)^2$
writing the subtracted part as
$f^{\prime\prime}\big(X(s)\big)\big(X(t)-X(s)\big)^2\int_0^1(1-\theta) d\theta$.