Question

A minimum value of ${\int }_0^{x}t{e}^{-t^2}dt$ is-

• A. $1$
• B. $2$
• C. $0$
• D. $3$

Answer 1

The correct option is C $0$

$f\left(x\right)={\int }_0^{x}t{e}^{-t^2}dt$
$f^{\prime }\left(x\right)=x{e}^{-x^2}$
For maxima or minima,
$f^{\prime }\left(x\right)=0$
$\Rightarrow x=0$
Now, $f^{\prime \prime }\left(x\right)=e^{-x^2}(1-2x^2)$
$f^{\prime \prime }\left(0\right)=1>0$
Hence, $f\left(x\right)$ has a minimum at $x=0$
Minimum value $=f\left(0\right)=0$