Question

Let $f:\left[ {0:2} \right] \to R$ be a function which is continuous on $\left[ {0,2} \right]$and is differentiable on $\left( {0,2} \right)$ with $f\left( 0 \right) = 1$.
Let $F\left( x \right) = \int\limits_0^{{x^2}} {f\left( {\sqrt t } \right)dt}$ for $x \in \left[ {0,2} \right]$. If $F'\left( x \right) = f'\left( x \right)$ for all $x \in \left( {0,2} \right)$, then $F\left( 2 \right)$ equals:
A. ${e^2} - 1$
B. ${e^4} - 1$
C. $e - 1$
D. ${e^4}$

Answer 1

In this problem we first use the given data to find the expression of $f\left( x \right)$ by integration and using the data $F'\left( x \right) = f'\left( x \right)$ and then we find the value of $c$. Then find the expression of $F\left( x \right)$ and substitute $x = 2$ to get the required solution.

Complete step-by-step answer:
Given that $F\left( x \right) = \int\limits_0^{{x^2}} {f\left( {\sqrt t } \right)dt}$ for $x \in \left[ {0,2} \right]$
Since the function $f\left( x \right)$ is continuous on $\left[ {0,2} \right]$ and is differentiable on$\left( {0,2} \right)$ differentiating the above expression w.r.t $x$ we have

$$\Rightarrow \dfrac{d}{{dx}}\left( {F\left( x \right)} \right) = \dfrac{d}{{dx}}\left( {\int\limits_0^{{x^2}} {f\left( {\sqrt t } \right)dt} } \right) \\\ \Rightarrow F'\left( x \right) = f\left( {\sqrt {{x^2}} } \right)\dfrac{d}{{dx}}\left( {{x^2}} \right) - f\left( 0 \right)\dfrac{d}{{dx}}\left( 0 \right) \\\ \Rightarrow F'\left( x \right) = f\left( {\left| x \right|} \right)\left( {2x} \right) - f\left( 0 \right) \times 0 \\\$$

Since $x \in \left[ {0,2} \right]$,$\left| x \right|$=$x$ then

$$\Rightarrow F'\left( x \right) = f\left( x \right)\left( {2x} \right) - f\left( 0 \right) \times 0 \\\ \Rightarrow F'\left( x \right) = 2f\left( x \right)x \\\$$

We are provided that $F'\left( x \right) = f'\left( x \right)$ then

$$\Rightarrow F'\left( x \right) = f'\left( x \right) = 2f\left( x \right)x \\\ \Rightarrow \dfrac{{f'\left( x \right)}}{{f\left( x \right)}} = 2x \\\$$

On integrating the above expression, we get

$$\Rightarrow \int {\dfrac{{f'\left( x \right)}}{{f\left( x \right)}}dx} = \int {2x} dx \\\ \Rightarrow \log \left( {f\left( x \right)} \right) = 2\left( {\dfrac{{{x^2}}}{2}} \right) + c \\\ \Rightarrow \log \left( {f\left( x \right)} \right) = {x^2} + c \\\$$

We have $f\left( 0 \right) = 1$ so we will proceed by putting $x = 0$ then above expression becomes

$$\Rightarrow \log \left( {f\left( 0 \right)} \right) = {\left( 0 \right)^2} + c \\\ \Rightarrow \log \left( 1 \right) = c \\\ \therefore c = 0 \\\$$

Then the expression becomes

$$\Rightarrow \log \left( {f\left( x \right)} \right) = {x^2} \\\ \therefore f\left( x \right) = {e^{{x^2}}} \\\$$

Then, $F\left( x \right) = \int_0^{{x^2}} {{e^x}} dx$
$\Rightarrow F\left( x \right) = {e^{{x^2}}} - 1$
Then putting $x = 2$

$$\Rightarrow F\left( 2 \right) = {e^2}^{^2} - 1 \\\ \Rightarrow F\left( 2 \right) = {e^4} - 1 \\\$$

\therefore $$$$F\left( 2 \right) = {e^4} - 1
Thus, the correct option is B. ${e^4} - 1$.

So, the correct answer is “Option B”.

Note: To solve these kinds of problems, always remember the formula $\dfrac{d}{{dx}}\left( {\int_{l.l}^{u.l} {f\left( x \right)dx} } \right) = f\left( {u.l} \right)\dfrac{d}{{dx}}\left( {u.l} \right) - f\left( {l.l} \right)\dfrac{d}{{dx}}\left( {l.l} \right)$ and $\log 1 = 0$. Here $f'\left( x \right)$ represents that it is the first derivative of $f\left( x \right)$ i.e., $\dfrac{d}{{dx}}\left( {f\left( x \right)} \right)$.