Proving an identity Given $a,b\in\mathbb{R}$ with $a < b$ and defining $F(z):=\int_0^z f(s) \, ds$ with $z \in \mathbb{R}$, how can one establish that $$F(a+b)=F(a)+f(a)b+ b^2\int_0^1 (1-s)f'(a+sb) \, ds,$$ which is part of a larger proof on page 508 of PDE Evans (2nd edition)?
Should I perform a change of variable to the second term on the RHS?
 A: First,
$$
F(a+b)-F(a)=\int_a^{a+b}f(t)\,dt.
$$
Then, integrate by parts,
$$
F(a+b)-F(a)=[(t-a-b)f(t)]_{a}^{a+b}-\int_a^{a+b}(t-a-b) f'(t)\,dt.
$$
Now, let $t=a+sb$,
$$
\begin{aligned}
F(a+b)-F(a)&=[(t-a-b)f(t)]_{a}^{a+b}+b^2\int_0^1 (1-s)f'(a+sb)\,ds\\
&=f(a)b+b^2\int_0^1 (1-s)f'(a+sb)\,ds.
\end{aligned}
$$
A: maybe useful to notice that if you take the author's relation:
$$
F(a+b)=F(a)+f(a)b+ b^2\int_0^1 (1-s)f'(a+sb) \, ds \
$$
and differentiate w.r.t $a$ you obtain
$$
f(a+b) = f(a)+ bf'(a) +b^2 \int_0^1 (1-s)f''(a+sb) \, ds \
$$
setting 
$$
I = b^2\int_0^1 (1-s)f''(a+sb) \, ds \
$$
we integrate by parts 
$$
I = [b(1-s)f'(a+sb)]_0^1 +b\int_0^1 f'(a+sb) \, ds \\
= -bf'(a) + f(a+b) -f(a)
$$
which is consistent. can you reverse the process?

Edit by @dragon:
Integration by parts establishes $$b^2 \int_0^1 (1-s) f''(a+sb) \, ds=-bf'(a)+f(a+b)-f(a).$$ So we have 
\begin{align}
f(a+b)&=f(a)+bf'(a)-bf'(a) + f(a+b) -f(a)\\&=f(a)+bf'(a)+b^2\int_0^1 (1-s) f''(a+sb) \, ds
\end{align}
Integrating both sides with respect to $a$, we obtain the textbook result $$F(a+b)=F(a)+bf(a)+b^2\int_0^1 (1-s) f'(a+sb) \, ds.$$
A: Taylor's formula with integral remainder: 
$$
F(a+b)=F(a) + F'(a) b + \int^{a+b}_a (a+b-t) F''(t)  \ dt
$$
Make a substitution $t=a+bs,$ and we get 
$$
F(a+b)=F(a) + F'(a) b + b^2\int^{1}_0 (1-s) F''(t) \ dt.
$$
Using $F(x)=\int^x_a f(w) \ dw,$ we get exactly the required result. 
