Question

Let $f:[0,2]\to R$ be a function which is continuous on [0,2] and is differentiable on (0,2) with f(0)=1.
Let $F\left(x\right)={\int }_0^{x^2}f\left(\sqrt{t}\right)dt,forxϵ[0,2],If{F}^{\prime }\left(x\right)=f^{\prime }\left(x\right),\forall xϵ(0,2),$ then F(2) equals

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

Answer 1

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


Given,
$F\left(x\right)={\int }_0^{x^2}f\left(\sqrt{t}\right)dt\Rightarrow F^{\prime }\left(x\right)=2xf\left(x\right)Also{F}^{\prime }\left(x\right)=f^{\prime }\left(x\right)\Rightarrow 2xf\left(x\right)=f^{\prime }\left(x\right)\Rightarrow \frac{f^{\prime }\left(x\right)}{f\left(x\right)}=2x\Rightarrow \int \frac{f^{\prime }\left(x\right)}{f\left(x\right)}=\int 2xdx\Rightarrow lnf\left(x\right)=x^2+c\Rightarrow f\left(x\right)=e^{x^2}+c\Rightarrow f\left(x\right)=K{e}^{x^2}[K=e^c]Now,f\left(0\right)=1\therefore 1=KHence,f\left(x\right)=e^{x^2}F\left(2\right)={\int }_0^4{e}^{t}dt=[e^t{]}_0^4{e}^4-1$