Question

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

Hint: Integrate the problem directly and take F(t) as the integration of Fâ€™(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,
$\therefore F(x)=\dfrac{1}{{{x}^{2}}}\left[ \int\limits_{4}^{x}{4{{t}^{2}}}dt-2\int\limits_{4}^{x}{F'(t)dt} \right]$â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦ (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,
$\therefore \int\limits_{4}^{x}{4{{t}^{2}}}dt=4\times \left[ \dfrac{{{x}^{3}}}{3}-\dfrac{{{4}^{3}}}{3} \right]$
$\therefore \int\limits_{4}^{x}{4{{t}^{2}}}dt=\dfrac{4}{3}\times \left[ {{x}^{3}}-{{4}^{3}} \right]$â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦. (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,
$\therefore \int\limits_{4}^{x}{F'(t)dt}=\left[ F(x)-F(4) \right]$â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.. (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 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦ (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 Fâ€™(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.