A minimum value of _0^x t e^ - t^2 dt is - Vidyadip

MCQ

A minimum value of $\int_0^x {t{e^{ - {t^2}}}} dt $ is

A
$1$
B
$2$
C
$3$
✓
$0$

✓

Answer

Correct option: D.

$0$

d
(d) $f(x) = \int_0^x {t{e^{ - {t^2}}}} dt \Rightarrow f'(x) = x{e^{ - {x^2}}} = 0 \Rightarrow x = 0$

$f''(x) = {e^{ - {x^2}}}(1 - 2{x^2});\;\;\;f''\,(0) = 1 > 0$

$\therefore $ Minimum value $f(0) = 0$.

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