Question

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

A

$1$

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B

$2$

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C

$3$

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D

$0$

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Solution

## 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}$$⇒$$\begin{array}{rcl}2tdt& =& du\end{array}$$⇒$ $\begin{array}{rcl}tdt& =& \frac{1}{2}du\end{array}$Now, $\begin{array}{rcl}f\left(x\right)& =& \frac{1}{2}{\int }_{0}^{{x}^{2}}{e}^{-u}du\end{array}$Integrate it with respect to $u$$\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}$ $\left[\mathbf{\because }\mathbf{\int }{\mathbit{e}}^{\mathbf{-}\mathbf{x}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{-}{\mathbit{e}}^{\mathbf{-}\mathbf{x}}\right]$Step 2. Find the minimum value, differentiate $f\left(x\right)$ and put $f\text{'}\left(x\right)=0$ $\begin{array}{rcl}\frac{d}{dx}\left(\frac{1}{2}\left[1-{e}^{-{x}^{2}}\right]\right)& =& 0\end{array}$ $⇒$$\begin{array}{rcl}0+{e}^{-{x}^{2}}\frac{d}{dx}\left({x}^{2}\right)& =& 0\end{array}$ $\left[\mathbf{\because }\frac{\mathbf{d}\left({e}^{-x}\right)}{\mathbf{d}\mathbf{x}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{-}{\mathbit{e}}^{\mathbf{-}\mathbf{x}}\right]$$⇒$ $\begin{array}{rcl}2x{e}^{-{x}^{2}}& =& 0\end{array}$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).

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