Answer
$3$
Work Step by Step
Let $w=t-2$.
Using the substitution method, $\int_2^{2+h}\sqrt{1+t^3}dt=\int_0^h\sqrt{1+(w+2)^3}dw$
It follows by the Fundamental Theorem of Calculus in Part 1 that $\frac{d}{dh}\int_0^h\sqrt{1+(w+2)^3}dw=\sqrt{1+(h+2)^3}=\sqrt{1+h^3+6h^2+12h+8}=\sqrt{h^3+6h^2+12h+9}$
Now, find the limit:
$\lim\limits_{h \to 0}\frac{1}{h}\int_2^{2+h}\sqrt{1+t^3}dt=\lim\limits_{h \to 0}\frac{\int_2^{2+h}\sqrt{1+t^3}dt}{h}$
$=\lim\limits_{h \to 0}\frac{\int_0^h\sqrt{1+(w+2)^3}dw}{h}$ (Apply the L'Hospital Rule)
$=\lim\limits_{h \to 0}\frac{{d}{dh}\int_0^h\sqrt{1+(w+2)^3}dw}{\frac{d}{dh}(h)}$
$=\lim\limits_{h \to 0}\frac{\sqrt{h^3+6h^2+12h+9}}{1}$
$=\lim\limits_{h \to 0}\sqrt{h^3+6h^2+12h+9}$ (Evaluate the limit by direct substitution)
$=\sqrt{0^3+6\cdot 0^2+12\cdot 0+9}$
$=\sqrt{9}$
$=3$
