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# If$F(x)=\dfrac{1}{{{x}^{2}}}\int\limits_{4}^{x}{\left( 4{{t}^{2}}-2F'(t) \right)}dt$, then F’(4) equals(a) $\dfrac{32}{9}$(b) $\dfrac{64}{3}$(c) $\dfrac{64}{9}$(D) $\dfrac{32}{3}$

Last updated date: 22nd Feb 2024
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Hint: Integrate the problem directly and take F(t) as the integration of F’(t). After finding the integration, differentiate it to find the solution.

We will write the given equation and will start integrating directly,
$F(x)=\dfrac{1}{{{x}^{2}}}\int\limits_{4}^{x}{\left( 4{{t}^{2}}-2F'(t) \right)}dt$
We can integrate both the terms separately,
$\therefore F(x)=\dfrac{1}{{{x}^{2}}}\left[ \int\limits_{4}^{x}{4{{t}^{2}}}dt-2\int\limits_{4}^{x}{F'(t)dt} \right]$………………………………………… (1)
Now, we will integrate the first term as shown below,
$\int\limits_{4}^{x}{4{{t}^{2}}}dt=4\int\limits_{4}^{x}{{{t}^{2}}}dt$
We can solve it further by using the formula given below,
Formula:
$\int{{{x}^{n}}}dx=\dfrac{{{x}^{n+1}}}{n+1}+c$
By using above formula we can write,
$\therefore \int\limits_{4}^{x}{4{{t}^{2}}}dt=4\times \left[ \dfrac{{{t}^{2+1}}}{2+1} \right]_{4}^{x}$
$\therefore \int\limits_{4}^{x}{4{{t}^{2}}}dt=4\times \left[ \dfrac{t3}{3} \right]_{4}^{x}$
We will substitute the limits to get the answer as shown below,
$\therefore \int\limits_{4}^{x}{4{{t}^{2}}}dt=4\times \left[ \dfrac{{{x}^{3}}}{3}-\dfrac{{{4}^{3}}}{3} \right]$
$\therefore \int\limits_{4}^{x}{4{{t}^{2}}}dt=\dfrac{4}{3}\times \left[ {{x}^{3}}-{{4}^{3}} \right]$…………………………………………. (2)
Also, to find$\int\limits_{4}^{x}{F'(t)dt}$ we should know the relation between integration and derivative,
If, $\dfrac{d}{dx}f(x)=f'(x)$ then $\int{f'(x)}=f(x)$
Therefore we can write $\int\limits_{4}^{x}{F'(t)dt}$ by using above formula as,
$\int\limits_{4}^{x}{F'(t)dt}=\left[ F(t) \right]_{4}^{x}$
We will substitute the limits to get the answer, as shown below,
$\therefore \int\limits_{4}^{x}{F'(t)dt}=\left[ F(x)-F(4) \right]$…………………………………….. (3)
Now put the values of equation (2) and (3) in equation (1)
$\therefore F(x)=\dfrac{1}{{{x}^{2}}}\left[ \int\limits_{4}^{x}{4{{t}^{2}}}dt-2\int\limits_{4}^{x}{F'(t)dt} \right]$
$\therefore F(x)=\dfrac{1}{{{x}^{2}}}\left[ \dfrac{4}{3}\times \left[ {{x}^{3}}-{{4}^{3}} \right]-2\left( F(x)-F(4) \right) \right]$
Multiply by $\dfrac{4}{3}$ in to the bracket we will get,
$\therefore F(x)=\dfrac{1}{{{x}^{2}}}\left[ \dfrac{4\times {{x}^{3}}}{3}-\dfrac{4\times {{4}^{3}}}{3}-2F(x)+2F(4) \right]$
Multiplying by $\dfrac{1}{{{x}^{2}}}$ in the bracket we will get,
$\therefore F(x)=\left[ \dfrac{4\times {{x}^{3}}}{3\times {{x}^{2}}}-\dfrac{4\times {{4}^{3}}}{3\times {{x}^{2}}}-\dfrac{2F(x)}{{{x}^{2}}}+\dfrac{2F(4)}{{{x}^{2}}} \right]$
$\therefore F(x)=\left[ \dfrac{4x}{3}-\dfrac{256}{3{{x}^{2}}}-\dfrac{2F(x)}{{{x}^{2}}}+\dfrac{2F(4)}{{{x}^{2}}} \right]$
To get the final answer we have to differentiate the above equation,
Therefore differentiating above equation w.r.t. x is given by,
$\therefore F'(x)=\dfrac{d}{dx}\left( \dfrac{4x}{3} \right)-\dfrac{d}{dx}\left( \dfrac{256}{3{{x}^{2}}} \right)-\dfrac{d}{dx}\left[ \dfrac{2F(x)}{{{x}^{2}}} \right]+\dfrac{d}{dx}\left[ \dfrac{2F(4)}{{{x}^{2}}} \right]$
We will take constants outside the derivative and rewrite the equation,
$\therefore F'(x)=\dfrac{4}{3}\dfrac{d}{dx}\left( x \right)-\dfrac{256}{3}\dfrac{d}{dx}\left( \dfrac{1}{{{x}^{2}}} \right)-2\dfrac{d}{dx}\left[ \dfrac{F(x)}{{{x}^{2}}} \right]+2F(4)\dfrac{d}{dx}\left[ \dfrac{1}{{{x}^{2}}} \right]$
By using the formula: $\dfrac{d}{dx}\left( x \right)=1$ we will get,
$\therefore F'(x)=\dfrac{4}{3}\times 1-\dfrac{256}{3}\dfrac{d}{dx}\left( \dfrac{1}{{{x}^{2}}} \right)-2\dfrac{d}{dx}\left[ \dfrac{F(x)}{{{x}^{2}}} \right]+2F(4)\dfrac{d}{dx}\left[ \dfrac{1}{{{x}^{2}}} \right]$
By using the formula: $\dfrac{d}{dx}\left( \dfrac{1}{{{x}^{n}}} \right)=\dfrac{-n}{{{x}^{n+1}}}$ we will get ………………………………… (4)
$\therefore F'(x)=\dfrac{4}{3}-\dfrac{256}{3}\times \dfrac{-2}{{{x}^{2+1}}}-2\dfrac{d}{dx}\left[ \dfrac{F(x)}{{{x}^{2}}} \right]+2F(4)\dfrac{d}{dx}\left[ \dfrac{1}{{{x}^{2}}} \right]$
By using formula of division rule i. e. $\dfrac{d}{dx}\left[ \dfrac{u}{v} \right]=\dfrac{v\dfrac{d}{dx}u-u\dfrac{d}{dx}v}{{{v}^{2}}}$
$\therefore F'(x)=\dfrac{4}{3}+\dfrac{256}{3}\times \dfrac{2}{{{x}^{3}}}-2\left[ \dfrac{{{x}^{2}}\dfrac{d}{dx}F(x)-F(x)\dfrac{d}{dx}({{x}^{2}})}{{{({{x}^{2}})}^{2}}} \right]+2F(4)\dfrac{d}{dx}\left[ \dfrac{1}{{{x}^{2}}} \right]$
We can get derivative of remaining terms by using formulae stated earlier in this problem,
$\therefore F'(x)=\dfrac{4}{3}+\dfrac{256}{3}\times \dfrac{2}{{{x}^{3}}}-\dfrac{2{{x}^{2}}\times F'(x)-2F(x)\times (2x)}{{{x}^{4}}}+2F(4)\left[ \dfrac{-2}{{{x}^{3}}} \right]$
Further algebraic simplification will give,
$\therefore F'(x)=\dfrac{4}{3}+\dfrac{256}{3}\times \dfrac{2}{{{x}^{3}}}-\dfrac{2{{x}^{2}}\times F'(x)}{{{x}^{4}}}+\dfrac{2F(x)\times (2x)}{{{x}^{4}}}-\dfrac{4F(4)}{{{x}^{3}}}$
$\therefore F'(x)=\dfrac{4}{3}+\dfrac{256}{3}\times \dfrac{2}{{{x}^{3}}}-\dfrac{2F'(x)}{{{x}^{2}}}+\dfrac{4F(x)}{{{x}^{3}}}-\dfrac{4F(4)}{{{x}^{3}}}$
$\therefore F'(x)=\dfrac{4}{3}+\dfrac{256}{3}\times \dfrac{2}{{{x}^{3}}}-\dfrac{2F'(x)}{{{x}^{2}}}$
Now as we have to find F’(4),
Put, x=4 in above equation,
$\therefore F'(4)=\dfrac{4}{3}+\dfrac{256}{3}\times \dfrac{2}{{{4}^{3}}}-\dfrac{2F'(4)}{{{4}^{2}}}$
$\therefore F'(4)+\dfrac{2F'(4)}{{{4}^{2}}}=\dfrac{4}{3}+\dfrac{256}{3}\times \dfrac{2}{64}$
$\therefore F'(4)+\dfrac{2F'(4)}{16}=\dfrac{4}{3}+\dfrac{4}{3}\times 2$
$\therefore F'(4)\left[ 1+\dfrac{2}{16} \right]=\dfrac{4}{3}+\dfrac{8}{3}$
$\therefore F'(4)\left[ \dfrac{16+2}{16} \right]=\dfrac{12}{3}$
$\therefore F'(4)\left[ \dfrac{18}{16} \right]=4$
$\therefore F'(4)=4\times \dfrac{16}{18}$
$\therefore F'(4)=2\times \dfrac{16}{9}$
$\therefore F'(4)=\dfrac{32}{9}$
Therefore option (a) is the correct answer.

Note: Do not use Leibniz Rule to get the derivative of the integral as it makes the problem lengthy and consumes your time.
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