Question

Let $f:[0,2]\to R$ be a function which is continuous on $[0,2]$ and is differentiable on $(0,2)$ with $f\left(0\right)=1.$ Let $F\left(x\right)=\underset{0}{\overset{x^2}{\int }}f\left(\sqrt{t}\right)dt,$ for $x\in [0,2]$. If $F^{\prime }\left(x\right)=f^{\prime }\left(x\right),\forall x\in (0,2),$ then $F\left(2\right)$ equals

• A. $e^2-1$
• B. $e^4-1$
• C. $e-1$
• D. $e^4$

Answer 1

The correct option is B $e^4-1$

We know that,
$F\left(x\right)=\underset{0}{\overset{x^2}{\int }}f\left(\sqrt{t}\right)dt$
$F\left(0\right)=0$
$F^{\prime }\left(x\right)=2xf\left(x\right)=f^{\prime }\left(x\right)\Rightarrow \int \frac{f^{\prime }\left(x\right)}{f\left(x\right)}dx=\int 2xdx\Rightarrow \ln f\left(x\right)=x^2+c$

Given, $f\left(0\right)=1\Rightarrow c=0$
$\Rightarrow f\left(x\right)=e^{x^2}$
So,
$F\left(x\right)=\underset{0}{\overset{x^2}{\int }}{e}^{t}dt\Rightarrow F\left(x\right)=[e^t{]}_0^{x^2}=e^{x^2}-1\Rightarrow F\left(2\right)=e^4-1$