Question

A minimum value of integral from $0$ to $xt{e}^{-t^2}dt$ is

• A. $1$
• B. $2$
• C. $3$
• D. $0$

Answer 1

The correct option is D

$0$

Explanation for the correct option:

Step 1. Finding the minimum value:

Given, $\begin{array}{rcl}f\left(x\right)& =& {\int }_0^{x}t{e}^{-t^2}dt\end{array}$
Substitute $t^2=u$ and differentiate it with respect to $t$

$\begin{array}{rcl}2t& =& \frac{du}{dt}\end{array}$

$\Rightarrow$$\begin{array}{rcl}2tdt& =& du\end{array}$

$\Rightarrow$ $\begin{array}{rcl}tdt& =& \frac{1}{2}du\end{array}$

Now, $\begin{array}{rcl}f\left(x\right)& =& \frac{1}{2}{\int }_0^{x^2}{e}^{-u}du\end{array}$

Integrate it with respect to $u$

$\begin{array}{rcl}f\left(x\right)& =& \frac{1}{2}{\overline{)\frac{e^{-u}}{-1}}}_0^{x^2}\\ & =& \frac{-1}{2}\left[e^{-x^2}-e^{-0}\right]\\ & =& \frac{-1}{2}\left[e^{-x^2}-1\right]\end{array}$ $\left[\mathbf{\because }\mathbf{\int }{\mathit{e}}^{\mathbf{-}\mathbf{x}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{-}{\mathit{e}}^{\mathbf{-}\mathbf{x}}\right]$

Step 2. Find the minimum value, differentiate $f\left(x\right)$ and put $f\text{'}\left(x\right)=0$

$\begin{array}{rcl}\frac{d}{dx}\left(\frac{1}{2}\left[1-e^{-x^2}\right]\right)& =& 0\end{array}$

$\Rightarrow$$\begin{array}{rcl}0+e^{-x^2}\frac{d}{dx}\left(x^2\right)& =& 0\end{array}$ $\left[\mathbf{\because }\frac{\mathbf{d}\left(e^{-x}\right)}{\mathbf{d}\mathbf{x}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{-}{\mathit{e}}^{\mathbf{-}\mathbf{x}}\right]$

$\Rightarrow$ $\begin{array}{rcl}2x{e}^{-x^2}& =& 0\end{array}$

if $x=0$, then $\begin{array}{rcl}2x{e}^{-x^2}& =& 0\end{array}$

$f\text{'}\left(x\right)$change sign from -ve to +ve ,so at $x=0,f\left(x\right)$ has local minima.

Hence, it is a point of minima

$\therefore$The minimum value is $\begin{array}{rcl}f\left(0\right)& =& {\int }_0^{0}t{e}^{-t^2}dt=0\end{array}$

Hence, The correct option is option (D).