Answer
\[ \lim_{h\rightarrow 0}\frac{\int_{2}^{2+h}\sqrt{1+t^3}\:dt}{h}=3\]
Work Step by Step
To find:-\[\lim_{h\rightarrow 0}\frac{\int_{2}^{2+h}\sqrt{1+t^3}\:dt}{h}\]
Which is $\frac{0}{0}$ form
We will use L' Hopital's rule which is \[\lim_{x\rightarrow 0}\frac{f(x)}{g(x)}=\lim_{x\rightarrow 0}\frac{f'(x)}{g'(x)}\;\;\;...(1)\]
Using (1)
\[\lim_{h\rightarrow 0}\frac{\int_{2}^{2+h}\sqrt{1+t^3}\:dt}{h}=\lim_{h\rightarrow 0}\frac{\frac{d}{dh}\int_{2}^{2+h}\sqrt{1+t^3}\:dt}{(h)'}\;\;\;...(2)\]
We will use the result \[\frac{d}{dx}\int_{f(x)}^{g(x)}F(t)\:dt=F(g(x))\cdot g'(x)-F(f(x))\cdot f'(x)\;\;\; ...(3)\]
Using (3) in (2)
\[\lim_{h\rightarrow 0}\frac{\int_{2}^{2+h}\sqrt{1+t^3}\:dt}{h}=\lim_{h\rightarrow 0}\frac{\sqrt{1+(2+h)^3}\cdot (1)}{1}\]
\[\Rightarrow \lim_{h\rightarrow 0}\frac{\int_{2}^{2+h}\sqrt{1+t^3}\:dt}{h}=\sqrt{1+(2)^3}=3\]
Hence \[ \lim_{h\rightarrow 0}\frac{\int_{2}^{2+h}\sqrt{1+t^3}\:dt}{h}=3\]
