$ F(x) = \int_0^2 \sin(x+l)^2\ dl$ Consider the function : $ F(x) =  \int_0^2 \sin(x+l)^2\ dl$, calculate $ \frac{dF(x)}{dx}|_{x=0}$ the derivative of $F(x)$ with respect to $x$ in zero.
Let $g(x) = \sin (x)$ and $h(x) = (x+l)^2$ then $F(x) = \int_0^2 g(h(x))\ dl$, so $F´(x) = g(h(x)) h´(x)$  then I can evaluate this in $x=0$ , so $F´(0) = \sin{l^2} 2l $ is that correct? Some help to calculate this please. 
 A: You need to differentiate under the integral sign: $$\frac{dF}{dx} = \frac{d}{dx} \int_0^2 \sin(x+l)^2 \, dl = \int_0^2 \frac{\partial}{\partial x} \sin (x+l)^2 \, dl = \int_0^2 \cos ((x+l)^2) \cdot 2(x+l) \, dl.$$ Therefore $$\frac{dF}{dx}\bigg\vert_{x=0} = \int_0^2 2l\cos(l^2) \, dl.$$ What you are trying to do is use the fundamental theorem of calculus, but you can't. The fundamental theorem of calculus asserts that $$\text{if } F(x) = \int_a^x f(t) \, dt \text{ then } \frac{dF}{dx} = f(x),$$ under reasonable conditions.
A: Considering $u=x+l$ :
$$F(x) = \int_x^{2+x} \sin(u)^2\ du$$
so$$\frac{dF}{dx} = \sin(2+x)^2-\sin(x)^2$$
hence$$\frac{dF}{dx}\bigg\vert_{x=0} = \sin(2)^2$$
A: If you can justify permuting the order of limit and integral then
\begin{align}
   \frac{dF(x)}{dx}
&= \lim_{h \to 0} \frac{1}{h}\left( \int_0^2 \sin(x + h+ \ell)^2 d\ell - 
   \int_0^2 \sin(x + \ell)^2 d\ell \right) \\
&= \lim_{h \to 0} \frac{1}{h} \int_0^2 (\sin(x + h+ \ell)^2 - \sin(x + \ell)^2) d\ell  \\
&= \int_0^2 \lim_{h \to 0} \frac{\sin(x + h+ \ell)^2 - \sin(x + \ell)^2}{h} d\ell  \\
&= \int_0^2 2(x + \ell) \cos(x + \ell)^2 d\ell \\
&= \int_x^{2+x} 2u \cos u^2 du \\
&= \left. \sin u^2 \right|^{2+x}_x \\
&= \sin(2+x)^2 - \sin x^2.
\end{align}
