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 $\underset{x\to 2}{lim}{\int }_6^{f\left(x\right)}\frac{4t^3}{x-2}dt$ equals?

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

Answer 1

The correct option is C $18$

${lim}_{x\to 2}\frac{1}{x-2}{\int }_6^{f\left(x\right)}4t^{3}dt=\frac{{\int }_6^{6}4t^{3}dt}{2-2}\frac{0}{0}$form

${lim}_{x\to 2}\frac{4{\left(f\left(x\right)\right)}^3\times f^{\prime }\left(x\right)-4{\left(6\right)}^3\times 0}{1}$

$=4{\left(f\left(2\right)\right)}^3\times f^{\prime }\left(2\right)$

$=4\times 6^3\times \frac{1}{48}=18$