Official solution
\[\mathop {\lim }\limits_{x \to 2} {{\int\limits_6^{f\left( x \right)} {2tdt} } \over {\left( {x - 2} \right)}}\]
This is \[{0 \over 0}\] form so we use L -- Hospital Rule.
= \[\mathop {\lim }\limits_{x \to 2} {{2f\left( x \right).f'\left( x \right) - 0} \over 1}\]
= \[{2f\left( 2 \right).f'\left( 2 \right)}\]
= 12f'(2)
Note : Newton - Leibnitz rule of differentiation under integration
\[{d \over {dx}}\int\limits_{\alpha \left( x \right)}^{\beta \left( x \right)} {f\left( u \right)} du\]
= \[f\left( {\beta \left( x \right)} \right)\beta '\left( x \right) - f\left( {\alpha \left( x \right)} \right)\alpha '\left( x \right)\]