Question

Let $f : [0,2] \rightarrow R$ be a function which is continuous on $[0,2]$ and is differentiable on $(0,2)$ with $f(0) = 1$. Let $F(x)= \int\limits_0^{x^2} \, f(\sqrt t)dt, \,$ for $\, x \in \, [0,2], if F'(x) \, = f'(x) , \forall$ $x \in \, (0,2) ,\,$ then $\, F(2)$ equals

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

Answer 1

Answer: (B)

$e^4 -1$

Answer 2

$F (0)=0$
$F ^{\prime}( x )=2 x f ( x )= f ( x )$
$f ( x )= e ^{ x ^{2}+ c }$
$f ( x )= e ^{ x ^{2}}(\because f (0)=1)$
$F ( x )=\int\limits_{0}^{ x ^{2}} e ^{ x } d x$
$F ( x )= e ^{ x ^{2}}-1(\because F (0)=0)$
$\Rightarrow F (2)= e ^{4}-1$

Therefore, the correct option is (B): $e^4 -1$