Does Constant factor rule in integration hold for functionals? The constant factor rule in integration states the following relation is valid

$$\int a f(x)dx=a \int f(x) dx$$ for all constants $a$(or $a$ that are constant functions of x, that is $\dfrac{da}{dx}=0$ )

My question is: Given a functional $$F[f(x)]=\int_{a}^{b} L(f(x))dx$$

Where $L$ here could be anything, like say $L(f(x))=f(x)$ or $L(f(x))=(f(x))^n$ or say $L(f(x))=\sin(f(x))$ and etc. 

Does the following relation hold?

$$\int_{a}^{b} F[f(x)] g(x)dx=F[f(x)] \int_{a}^{b} g(x) dx$$ 

It seems that it holds(because I saw it used in a proof of the product rule for functional derivatives). However consider the following: let's evaluate $\dfrac{dF}{dx}$ and check if it's zero or not.

$$\dfrac{dF}{dx}=\dfrac{d}{dx}\int_{a}^{b} L(f(x))dx$$

By the fundamental theorem of calculus we have

$$\dfrac{dF}{dx}= L(f(b))-L(f(a))=\text{constant}$$

Which is not zero. therefore $F$ is not a constant function of $x$ therefore the  Constant factor rule in integration does not hold for functionals.

Another way to put it: why cannot we think of the functional $F$ just like an ordinary function of $f$(like $g(f(x))$)? and since $f$ depends of $x$ therefore, $F$ will have implicit dependency on $x$, hence its total derivative is not zero.

So does it hold or not?
So My fault was in this step $$\dfrac{dF}{dx}=\dfrac{d}{dx}\int_{a}^{b} L(f(x))dx$$ When evaluating this expression, I thought that one can exchange the derivative and the integral if $L(f(x))$ is a continuous function, but it turns out I cannot.And since this is a definite integral, it'll evaluate to a constant, and its derivative is zero, therefore $$\dfrac{dF}{dx}=0$$

Another way to put this is that, Leibniz integral rule states

$$\dfrac{d}{dx}\int_{u(x)}^{v(t)}f(t)dt=f(v(x))v'(x)-f(u(t))u'(x)$$

In our case $v(x)=b$ and $u(x)=a$ where they're constants, therefore $$v'(x)=u'(x)=0$$

Therefore $$\dfrac{dF}{dx}=0$$
It doesn't hold in general. It holds for all functionals $L$ such that $$L(a f(x)) = a L(f(x))$$