The minimum value of int_0^x {t{e^{ - {t^2}}} | vedclass

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(D) Let $f(x) = \int_0^x {t{e^{ - {t^2}}}} dt$.By the Fundamental Theorem of Calculus,$f'(x) = x{e^{ - {x^2}}}$.To find the critical points,set $f'(x)

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(D) Let $f(x) = \int_0^x {t{e^{ - {t^2}}}} dt$.By the Fundamental Theorem of Calculus,$f'(x) = x{e^{ - {x^2}}}$.To find the critical points,set $f'(x) = 0$,which gives $x{e^{ - {x^2}}} = 0$. Since ${e^{ - {x^2}}} 
eq 0$ for any real $x$,we have $x = 0$.Now,find the second derivative: $f''(x) = \frac{d}{dx}(x{e^{ - {x^2}}}) = {e^{ - {x^2}}} + x({e^{ - {x^2}}}(-2x)) = {e^{ - {x^2}}}(1 - 2{x^2})$.Evaluating at $x = 0$: $f''(0) = {e^0}(1 - 0) = 1$.Since $f''(0) > 0$,the function has a local minimum at $x = 0$.The minimum value is $f(0) = \int_0^0 {t{e^{ - {t^2}}}} dt = 0$.

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

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