Evaluate the following limit: $\lim\limits_{h\to 0} \frac {1}{h} \int_2^{2+h}\sin(x^2) \, dx $ Evaluate the following limit:
$$\lim_{h\to 0} \frac {1}{h} \int_2^{2+h} \sin(x^2) \, dx $$
Would this be correct? 
\begin{align}
& \lim_{h\to 0} \frac {1}{h} \int_2^{2+h} \sin(x^2) \, dx \\[10pt]
= {} & \lim_{h\to0} \frac{ \int_0^{2+h} \sin(x^2) \, dx -\int_0^2 \sin(x^2) \, dx}{h}\\[10pt]
= {} &\lim_{h\to0} \frac{f(2+h)-f(2)} h = f'(2)
= {} \sin(2^2) = \sin 4
\end{align}
where
$$f(u) = \int_0^u \sin(x^2)\, dx.$$
Or must I define $f(u)$ as follows?
$$f(u) = \int_2^u \sin(x^2)\, dx.$$

\begin{align}
& \lim_{h\to 0} \frac {1}{h} \int_2^{2+h} \sin(x^2) \, dx \\[10pt]
= {} & \lim_{h\to0} \frac{ \int_2^{2+h} \cdots -\int_2^2 \cdots }{h} \\[10pt]
= {} &\lim_{h\to0} \frac{f(2+h)-f(2)} h = f'(2)
\end{align}
where
$$
f(u) = \int_2^u \sin(x^2)\, dx.
$$
So use the fundamental theorem of calculus to find $f'(2).$ (You get $\sin(2^2).$)
