Question

Let $f:\left[0,2\right]\to R$ 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 $F\left(x\right)=\underset{0}{\overset{x^2}{\int }}f\left(\sqrt{t}\right)dt,$ for $x\in \left[0,2\right]$, if $F^{\prime }\left(x\right)=f^{\prime }\left(x\right),\forall x\in \left(0,2\right),$ then $F\left(2\right)$ equals to

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

Answer 1

Answer: (B)

$e^4-1$

From Newton-Leibnitz's formula,

$\frac{d}{dx}\left[{\int }_{\varphi \left(x\right)}^{\psi \left(x\right)}f\left(t\right)dt\right]=f\left\{\psi \right(x\left)\right\}\left\{\frac{d}{dx}\psi \right(x\left)\right\}-f\left\{\varphi \right(x\left)\right\}\left\{\frac{d}{dx}\varphi \right(x\left)\right\}$

Given that,

$F\left(x\right)={\int }_0^{x^2}f\left(\sqrt{t}\right)dt$

$\therefore F^{\prime }\left(x\right)=f\left(x\right)\cdot \frac{d}{dx}\left(x^2\right)-f\left(0\right)\cdot \frac{d}{dx}\left(0\right)$

$\Rightarrow F^{\prime }\left(x\right)=2xf\left(x\right)...\left(i\right)$

Also given that,

$F^{\prime }\left(x\right)=f^{\prime }\left(x\right)$

$\Rightarrow 2xf\left(x\right)=f^{\prime }\left(x\right)$ [ from $\left(i\right)$]

$\Rightarrow \frac{f^{\prime }\left(x\right)}{f\left(x\right)}=2x$

Integrating both sides w.r.t. $x$,

$\Rightarrow \int \frac{f^{\prime }\left(x\right)}{f\left(x\right)}dx=\int 2xdx$

$\Rightarrow \ln \left\{f\right(x\left)\right\}=x^2+c$

$\Rightarrow f\left(x\right)=e^{x^2+c}$

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

Now, $f\left(0\right)=1$

$\Rightarrow k=1$

Hence,$f\left(x\right)=e^{x^2}$

$\therefore F\left(2\right)={\int }_0^4{e}^{t}dt=[e^t{]}_0^4=e^4-1$