Question

Let $f:\left( {0, + \infty } \right) \to R $ and $F\left( x \right) = $ integral from $0$ to $x{\text{ }}f\left( t \right)dt$. If$F\left( {{x^2}} \right) = {x^2}\left( {1 + x} \right)$, then $f\left( 4 \right) = $?
A. $\dfrac{5}{4}$
B. $7$
C. $4$
D. $2$

Answer 1

Hint: Differentiation is the rate at which one quantity changes in relation to another, while integration is the process of determining a function's antiderivative.
To determine the value of $f(4)$, we need to differentiate the given function $F\left( {{x^2}} \right) = {x^2}\left( {1 + x} \right)$ with respect to $x$ then substitute ${x^2} = t$. After that, use the Leibnitz theorem on the obtained function.

Complete step by step Solution:
In the question, the function $F\left( {{x^2}} \right) = {x^2}\left( {1 + x} \right)$ is given,

Differentiate the given function with respect to $x$, then:
$\dfrac{d}{{dx}}(F\left( {{x^2}} \right)) = \dfrac{d}{{dx}}({x^2}\left( {1 + x} \right)) \\$
   $\Rightarrow 2xF\prime ({x^2}) = 2x(1 + x) + {x^2} \\$
   $\Rightarrow F\prime ({x^2}) = (1 + x) + \dfrac{x}{2} \\$

Substitute ${x^2} = t$in the above equation, if ${x^2} = t$then we have $x = \sqrt t $:
$F\prime (t) = (1 + \sqrt t ) + \dfrac{{\sqrt t }}{2}\,\,\,\,\,.....(1)$

According to the given information, $f:\left( {0, + \infty } \right) \to R$and $F\left( x \right) = $integral from $0$to $x{\text{ }}f\left( t \right)dt$, so:
$F(x) = \int\limits_0^x {f(t)dt} $

Now using the Leibnitz theorem, then:
$F\prime (x) = 1 \cdot f(x) + 0 \\$
 $\Rightarrow F\prime (x) = f(x) \\$

Substitute the value from $(1)$in the obtained function, we obtain:
$f(x) = (1 + \sqrt x ) + \dfrac{{\sqrt x }}{2}$

Put $x = 4$in the function $f(x) = (1 + \sqrt x ) + \dfrac{{\sqrt x }}{2}$to determine the value of $f(4)$, $f(4) = (1 + \sqrt 4 ) + \dfrac{{\sqrt 4 }}{2} \\$
$\Rightarrow f(4) = (1 + 2) + \dfrac{2}{2} \\$
$\Rightarrow f(4) = 3 + 1 \\$
$\Rightarrow f(4) = 4 \\$
Therefore, the value of $f(4)$is $4$.

Hence, the correct option is C.

Note: The Leibnitz theorem is essentially the Leibnitz rule defined for the derivative of the antiderivative, it should be noted. As per the rule, the derivative on ${n^{th}}$ order of the product of two functions works as a connection between integration and differentiation.