Let $f: R \rightarrow R$ be a differentiable function having $f(2)=6, f'(2)=\frac{1}{48}$. Then, $\lim\limits_{x \rightarrow 2} \int\limits_6^{f(x)} \frac{4 t^3}{x-2} d t$, equals

18

We have,

$I=\lim\limits_{x \rightarrow 2} \int\limits_6^{f(x)} \frac{4 t^3}{x-2} d t=\lim\limits_{x \rightarrow 2} \frac{\int\limits_6^{f(x)} 4 t^3 d t}{x-2}$

$\Rightarrow I=\lim\limits_{x \rightarrow 2} \frac{4\{f(x)\}^3 f'(x)}{1}$                          [Using L' Hospitals rule]

$\Rightarrow I=4\{f(2)\}^3 f'(2)=4 \times 216 \times \frac{1}{48}=18$