Prove that $f(x)=\int_{-\infty}^{x} e^{\frac{-t^2}{2}} dt$ for $x\in\mathbb{R}$ is uniformly continuous
Okay so I am unsure how to start this! I can prove a function is uniformly continuous but I'm not sure how to adapt that to an integral?
2026-04-01 20:30:24.1775075424
Proving an integral is uniformly continuous: $f(x)=\int_{-\infty}^{x} e^{\frac{-t^2}{2}} dt$
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We have $$ f'(x)= e^{\frac{-x^2}{2}},\quad\forall x\in\mathbb{R}, $$ and $$ \left|f'(x)\right|=\left| e^{\frac{-x^2}{2}}\right|\leq1,\quad\forall x\in\mathbb{R}, $$ hence by the mean value theorem we get $$|f(x)-f(y)|\leq |x-y|\quad \forall x,y\in\mathbb{R}$$ so $f$ is a lipschitzian function on $\mathbb{R}$ and therefore it is uniformly continuous on $\mathbb{R}$.