Discussion Forum

Let f:[0,2]→  R  be a function which is continuous on [0,2] and is differentiable on (0,2)  with f(0)=1.Let   $F(x)=\int_{0}^{x^{2}} f(\sqrt{t}) dt, for x\in [0,2]$ . If  F'(x)= f '(x) for all  x ε (0,2) , then F(2) equal to

Answer:

Explanation:

 Plan    Newton - Leibnitz.s formula

   $\frac{d}{dx}\left[\int_{\phi(X)}^{\psi (X)} f(t) dt \right]= f\left\{\psi(X)\right\}\left\{\frac{d}{dx}\psi(X)\right\}-f(\phi(X))\left\{\frac{d}{dx}\phi(X)\right\}$

 Given, $F(X)=\int_{0}^{x^{2}} f(\sqrt{t})dt$

 $\therefore$  F'(x)= 2x f(x)

 Also, F '(x)= f' (x)

$\Rightarrow$    $2x f(x)=f'(x)\Rightarrow \frac{f'(x)}{f(x)}=2x$

$\Rightarrow$  $\int_{}^{} \frac{f '(x)}{f(x)}dx=\int_{}^{} 2x dx$

$\Rightarrow$      ln f(x)= x²+ c

$\Rightarrow$     $f(x)= e^{x^{2}}+c$

$\Rightarrow$   $f(x)= K e^{x^{2}}$             $[K=e^{c}]$

 Now, f(0)=1

                               1=K

 Hence     $f(x)=  e^{x^{2}}$

       $F(2)=\int_{0}^{4} e^{t} dt$

    $=[e^{t}]_{0}^{4}=e^{4}-1$