What is the minimum value of intlimits_0^x {t | vedclass

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(C) Let $f(x) = \int\limits_0^x {t{e^{ - {t^2}}}} dt$. By the Fundamental Theorem of Calculus,the derivative is $f'(x) = x{e^{ - {x^2}}}$. Setting $f'

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(C) Let $f(x) = \int\limits_0^x {t{e^{ - {t^2}}}} dt$. By the Fundamental Theorem of Calculus,the derivative is $f'(x) = x{e^{ - {x^2}}}$. Setting $f'(x) = 0$,we get $x{e^{ - {x^2}}} = 0$. Since ${e^{ - {x^2}}} 
eq 0$ for all real $x$,we have $x = 0$. Now,find the second derivative: $f''(x) = {e^{ - {x^2}}} - 2{x^2}{e^{ - {x^2}}} = {e^{ - {x^2}}}(1 - 2{x^2})$. Evaluating at $x = 0$,$f''(0) = {e^0}(1 - 0) = 1 > 0$. Since the second derivative is positive at $x = 0$,the function has a local minimum at $x = 0$. The minimum value is $f(0) = \int\limits_0^0 {t{e^{ - {t^2}}}} dt = 0$.

What is the minimum value of $\int\limits_0^x {t{e^{ - {t^2}}}} dt$?

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