Let F:R\rightarrow R  be a differentiable function having f(2)=6,f'(2)=\left ( \frac{1}{48} \right ).\, Then\; \lim_{x\rightarrow 2}\, \int_{6}^{f(x)}\frac{4t^{3}}{x-2}\; dt\;  equals

  • Option 1)

    36

  • Option 2)

    24

  • Option 3)

    18

  • Option 4)

    12

 

Answers (1)

As we learnt in 

NEWTON LEIBNITZ THEOREM -

\frac {d}{dt}\left ( \int_{f(t)}^{\phi (t))}F(x)dx \right )=F(\phi(t))\phi^{'}(t)-F(f(t))f^{'}(t)

-

 

 f (2) = 6 ; f ' (2) = 1/48

\lim_{x\rightarrow 2}\int_{6}^{f(x)}\frac{4t^{3}dt}{x-2}= \lim_{x\rightarrow 2}\frac{4(f(x))^3 \times f'(x)}{1}

\Rightarrow 4(f(2))^3 \times f'(2)

\Rightarrow 4 \times 6^3 \times \frac{1}{48} = \frac{4\times216}{48}=18

 


Option 1)

36

This is incorrect option

Option 2)

24

This is incorrect option

Option 3)

18

This is correct option

Option 4)

12

This is incorrect option

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