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. 18
• B. 12
• C. 36
• D. 24

Answer 1

The correct option is A 18

$\underset{x\to 2}{lim}{\int }_6^{f\left(x\right)}\frac{4t^3}{x-2}dt=\underset{x\to 2}{lim}\frac{{\int }_6^{f\left(x\right)}4t^{3}dt}{x-2}=\underset{x\to 2}{lim}\frac{4f(x{)}^3}{1}{f}^{\prime }\left(x\right)=4{f\left(2\right)}^3{f}^{\prime }\left(2\right)=4\times (6{)}^3\times \frac{1}{48}=18$