Question

Let $f:R\to R$ be a differentiable function having $f\left(2\right)=6,$ $f^{{}^{\prime }}\left(2\right)=\left(\frac{1}{48}\right)$. Then, $\underset{x\to 2}{lim}{\int }_6^{f\left(x\right)}\frac{4t^3}{x-2}dt$ is equal to

• A. $18$
• B. $12$
• C. $36$
• D. $24$

Answer 1

The correct option is A $18$

The value of $\underset{x\to 2}{lim}{\int }_6^{f\left(x\right)}\frac{4t^3}{x-2}dt$ is
$=\underset{x\to 2}{lim}\frac{4{\left\{f\left(x\right)\right\}}^3}{1}\cdot f^{{}^{\prime }}\left(x\right)$ .....(By Leibnitz rule)
$=4{\left\{f\left(2\right)\right\}}^3\cdot f^{{}^{\prime }}\left(2\right)$
$=4\times {\left(6\right)}^3\cdot \frac{1}{48}=18$