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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

Posted by

Sabhrant Ambastha

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