Let f : R \[ \to \] R be a continuously differentiable function such that f(2) = 6 and f'(2) = \[{1 \over {48}}\]. If \[\int\limits_6^{f\left( x \right)} {4{t^3}} dt\] = (x - 2)g(x), then \[\mathop {\lim }\limits_{x \to 2} g\left( x \right)\] is equal to :
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Official solution
Given \[\int\limits_6^{f(x)} {4{x^3}dx} = g(x).(x - 2)\]
\[ \Rightarrow g(x) = \] \[{{\int\limits_0^{f\left( x \right)} {4{x^3}dx} } \over {x - 2}}\]
\[ \therefore \] \[\mathop {\lim }\limits_{x \to 2} g(x) = \] x→2limx−20∫f(x)4x3dxAt x = 2 this limit is in{0 \over 0}\[ form.
So we can use L'Hopital's rule. Use leibniz intgral rule to differentiate the integration.
\] \Rightarrow \mathop {\lim }\limits_{x \to 2} {{4{f^3}(x).f'(x)} \over 1} = 4 \times {6^3} \times {1 \over {48}} = 18