Differentiate $f \mapsto \int \limits_{f(0)}^{f(1)} f(t) dt$ Let $U=\{f \in \mathcal{C}([0,1],\mathbb{R}): 0<f(0)<1, 0<f(1)<1 \} \subset \mathcal{C}([0,1],\mathbb{R})$. Consider a function $F:U \to \mathbb{R}$ given by $f \mapsto \int \limits_{f(0)}^{f(1)} f(t) \ dt$. I want to find the derivative of this function.
By definition the derivative at point $f$ (in our case this is a function) is such linear function $L:\mathcal{C}([0,1],\mathbb{R}) \to \mathbb{R}$ such that
$$\forall \varepsilon >0 \ \exists r>0 : (||h||<r, f+h \in U) \Longrightarrow (|F(f+h)-F(f)-L(h)|\leq \varepsilon||h||)$$
When we substitute $F$ by the integral we have:
$$\left|\ \int \limits_{f(0)}^{f(1)} h(t) dt - L(h) \ \right| \leq \varepsilon||h||$$
And here I'm thinking about using something like integration by substitution
and I will have
$$\left|\ \int \limits_{0}^{1} f(h(t))h'(t) dt - L(h) \ \right| \leq \varepsilon||h||,$$
however there is no guarantee that $h$ is differentable.
How to find such $L$?
Thanks in advance
 A: Fix $f\in U$ and consider $h$ with $\|h\|$ small. Then
$$ F(f+h) = \int_{f(0) + h(0)}^{f(1) + h(1)} (f+h)$$ $$\tag 1 = \int_{f(0)}^{f(1)}(f+h) +\int_{f(1)}^{f(1) + h(1)}(f+h) - \int_{f(0)}^{f(0) + h(0)}(f+h).$$
In $(1)$ we see three integrals, let's call them $I_1,I_2, I_3.$ We have
$$\tag 2 I_1 = F(f) + \int_{f(0)}^{f(1)}h,$$
which is good because we need to see $F(f)$ isolated, and the integral in $(2)$ is linear in $h.$ Continuing,
$$\tag 3 I_2 = \int_{f(1)}^{f(1) + h(1)}f + \int_{f(1)}^{f(1) + h(1)}h.$$
Let's think about $(3):$ As $\|h\| \to 0,$ we expect the first integral to be close to $f(f(1))h(1)$ by the continuity of $f$ at $f(1).$ In fact it will equal that at the expense of a $o(\|h\|)$ term. The second integral in $(3)$ is clearly $o(\|h\|).$ Thus $(3)$ equals $f(f(1))h(1) +o(\|h\|).$ Similarly, $I_3 = f(f(0))h(0) + o(\|h\|).$
Putting everything together gives
$$F(f+h)= F(f) + \int_{f(0)}^{f(1)}h + f(f(1))h(1) - f(f(0))h(0) + o(\|h\|).$$
Thus $DF(f)(h) = \int_{f(0)}^{f(1)}h + f(f(1))h(1) - f(f(0))h(0).$
Thanks to Daniel Fischer for noticing a mistake, now corrected.
