Integration with limits and options. I found this exercise in an old exam but I don't know how to attack it because is a limit of an integration and I don't know if the limit affects the process of the integral or it makes it easier. The exercise is this:

The value of the limit 
  $$
\lim_{h\to 0}\frac{1}{h}\int^{2+h}_2\sqrt{1+t^3}dt
$$
  Is:
a) $0$
b) $1$
c) $\sqrt{2}$
d) $2$
e) $3$

I want to know two things: 
1) If there is a method of how to solve limit-integral problems.
2) An explained solution of this example to at least try to understand how to solve exercises of the same style. 
 A: As $h \rightarrow 0$, we have $\displaystyle \int_2^{2+h} \sqrt{1+t^3} \ dt \rightarrow 0$.  Thus, we arrive at a $0/0$ indeterminant form when trying to evaluate the limit, so we can apply L'Hopital's rule.
The hardest part about doing this is evaluating $\displaystyle \frac{d}{dh}\int_2^{2+h} \sqrt{1 + t^3} \ dt$.  For a continuous function $f,$ the fundamental theorem tells us that $\displaystyle \frac{d}{dx} \int_a^x f(t) \ dt = f(x)$.  To get our integral into this form, we can simply shift the function to the left $2$ units and adjust the bounds of integration accordingly:
$$\displaystyle \frac{d}{dh}\int_2^{2+h} \sqrt{1 + t^3} \ dt = \frac{d}{dh} \int_0^h \sqrt{1 + (t+2)^3} \ dt = \sqrt{1+(h+2)^3}$$
A: Use L'Hopital's rule as follows:
$$\lim_{h \to 0} \frac{\int_{2}^{2 + h} \sqrt{1 + t^3}dt}{h}$$
$$= \lim_{h \to 0} \frac{\sqrt{1 + (2 + h)^3}}{1}$$
$$= \boxed{3}.$$
And we are finished. Hope this helped!
A: By the Mean Value Theorem for Definite Integrals, https://en.wikipedia.org/wiki/Mean_value_theorem#First_Mean_Value_Theorem_for_Definite_Integrals,
$\displaystyle\int_2^{2+h}\sqrt{1+t^3}dt=\left(\sqrt{1+c^3}\right)h$ for some c between 2 and $2+h$, 
so $\displaystyle\lim_{h\to 0}\frac{1}{h}\int^{2+h}_2\sqrt{1+t^3}dt=\lim_{c\to 2}\sqrt{1+c^3}=\sqrt{9}=3$.

$\textbf{Alternate solution:}$
Let $\displaystyle G(x)=\int_2^x \sqrt{1+t^3}dt,\;$ so $G^{\prime}(x)=\sqrt{1+x^3}$ by the Fundamental Theorem of Calculus.
Then $\displaystyle\lim_{h\to 0}\frac{1}{h}\int^{2+h}_2\sqrt{1+t^3}dt=\lim_{h\to 0}\frac{G(2+h)-G(2)}{h}=G^{\prime}(2)=\sqrt{1+2^3}=3$
A: You should recognize that for well behaved $f(x)$ and fixed $x$:
 $$\frac{1}{h}\int^{x+h}_x f(t)dt=\frac{1}{h}\left[f(x)h+R_h(x) \right]
$$
where $\lim_{h\to 0}R_h(x)/h=0$. So:
$$
\lim_{h\to 0}\frac{1}{h}\int^{x+h}_x f(t)dt=f(x)
$$
An alternate way of showing this is to observe that for a well behaved function on an interval containing $[x,x+h]$:
$$
h \times \min_{t\in[x,x+h]}f(t)\le\int^{x+h}_x f(t)dt\le h \times \max_{t\in[x,x+h]}f(t)
$$
and when $h\to 0$ we have $\min_{t\in[x,x+h]}f(t)\to \max_{t\in[x,x+h]}f(t)\to f(x)$.
Hence as $f(x)=\sqrt{1+t^3}$ is well behaved around $t=2$:
$$
\lim_{h\to 0}\frac{1}{h}\int^{2+h}_2\sqrt{1+t^3}dt=\left.\sqrt{1+t^3}\right|_{t=2}=3
$$
A: You need to use Fundamental Theorem of Calculus. Let $$F(x) = \int_{2}^{x}\sqrt{1 + t^{3}}\,dt = \int_{2}^{x}f(t)\,dt$$ where $f(x) = \sqrt{1 + x^{3}}$ and we need to calculate the limit $$\lim_{h \to 0}\frac{F(2 + h)}{h}$$ Note that $F(2) = 0$ and hence $$\lim_{h \to 0}\frac{F(2 + h)}{h} = \lim_{h \to 0}\frac{F(2 + h) - F(2)}{h} = F'(2)$$ provided the limit exists.
By Fundamental Theorem of Calculus we know that $F'(x)$ exists if the function $f(x) = \sqrt{1 + x^{3}}$ is continuous at $x$ and then $F'(x) = f(x)$. Thus $F'(2) = f(2) = 3$.
The use of L'Hospital's Rule is unnecessary and is a roundabout way to evaluate this limit.
