Let f(x) = intlimits_0^x {{e^{ - {t^2}}}dt} | vedclass

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(D) Consider the function $g(t) = e^{-t^2}$. For $t \in (0, x)$,the function $g(t)$ is strictly decreasing because $g'(t) = -2t e^{-t^2}

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$f(x) > \frac{x}{2}\left( {1 + {e^{ - {x^2}}}} \right)$

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(D) Consider the function $g(t) = e^{-t^2}$. For $t \in (0, x)$,the function $g(t)$ is strictly decreasing because $g'(t) = -2t e^{-t^2}  0$.Since $g(t)$ is a strictly decreasing concave function on $(0, x)$,the area under the curve $y = e^{-t^2}$ from $0$ to $x$ is greater than the area of the trapezoid formed by the points $(0, 0), (x, 0), (x, e^{-x^2}),$ and $(0, 1)$.The area of this trapezoid is given by $\frac{1}{2} 	imes (	ext{sum of parallel sides}) 	imes (	ext{height}) = \frac{1}{2} (1 + e^{-x^2}) 	imes x$.Therefore,$f(x) = \int_0^x e^{-t^2} dt > \frac{x}{2} (1 + e^{-x^2})$.

Let $f(x) = \int\limits_0^x {{e^{ - {t^2}}}dt} $ for all $x > 0$. Then for all $x > 0$:

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