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I need to present a solution to the following problem. I will be glad if someone will check my arguments, since I'm not sure they are correct: Let $\phi(x) =\int_{x}^{x+1} e^{-t^2} dt $ . Where does the function $\phi$ is defined? Prove that $\phi$ doesn't have an intersection with the $x$ axis.

My attempt: I wrote $\phi(x)= F(x+1) - F(x)$ ,where $F(x)= \int_0^{x} e^{-t^2} dt $. Then, since the function $e^{-x^2} $ is continous in all $\mathbb{R} $ , we get that $F(x)$ is defined for every $x$ , and in particular $\phi$ is defined for every $x$ .

As for the second question: I was able to prove it using Rolle's theorem, but I am not sure about the following (possible?) way: $e^{-x^2}$ is continous, and positive. Thus $\int_a^b e^{-x^2} dx $ is strictly positive for every $[a,b]$ and in particular the function $\phi(x)$ is positive. Is this argument correct?

Thanks in advance everyone!

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up vote 0 down vote accepted

Both your arguments are correct while I prefer this for the second:

As for any $x\in \mathbb{R}$, $[x,x+1]$ ($x+1>x$) is a closed non empty interval and $e^{-x^2}>0$, $\phi(x)>0$ by the monotonicity of the integral. Note that the fact that $[x,x+1]$ is non-empty is very important and should be emphasised. Consider for example: $f(x)=\int_{0}^xe^{-t^2}dt$ Can the same argument be made in this case?

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Great ! Thanks a lot! Got it – theMissingIngredient Jan 6 '13 at 9:28

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