Proof that $\frac{2}{e}<\int_{-1}^1e^{-x^2}\, dx<2$ I have to prove that $\frac{2}{e}<\int_{-1}^1e^{-x^2}\, dx<2$.
To prove this I have thought to use two methods:\

*

*In $[-1,1]$ I have $\frac{1}{e}=e^{-1}\leq e^{-x^2}\leq e^{0}=1\implies \frac{2}{e}\int_1^{-1} \frac{1}{e}\, dt\leq \int_{-1}^1e^{-x^2}\, dx\leq \int_1^{-1}1\, dt=2 $.
$\color{red}{First\, doubt:}$ I have to obtain inequalities as $<$ and not $\leq$, so how can I pass from $$\frac{2}{e}\int_1^{-1} \frac{1}{e}\, dt\leq \int_{-1}^1e^{-x^2}\, dx\leq \int_1^{-1}1\, dt=2$$ to $$\frac{2}{e}\int_1^{-1} \frac{1}{e}\, dt< \int_{-1}^1e^{-x^2}\, dx< \int_1^{-1}1\, dt=2$$?

*Alternatively I have tought to apply the meanvalue theorem and so: $\exists c\in[-1,1]:$ $\int_{-1}^1f(t)\, dt=f(c)\cdot 2=2e^{-c^2}$. Hence:
$$\frac{2}{e}<2e^{-c^2}<2\text{ in}\,[-1,1]\iff c\neq 0$$
$\color{red}{Second\, doubt:}$ Now how can I proceede? I have to prove that $c\neq 0$?

 A: For all $x\in (-1,1)$, we have $e^{-1}< e^{-x^2}$, so the integral will preserve the strict inequality. As long as we have weak inequality everywhere, and a strict inequality on a set of positive measure (for example if the functions are continuous like in this case, and you have a strict inequality at one point, then it is enough to guarantee the integral retains the strict inequality). Here's the formal statement in the setting of Riemann-integration:

Let $f:[a,b]\to\Bbb{R}$ be Riemann integrable, and suppose $f\geq 0$. Then, $\int_a^bf \geq 0$. If in addition, there is a point $x_0\in [a,b]$ where $f(x_0)>0$ and such that $f$ is continuous at $x_0$, then $\int_a^bf>0$.

I highly suggest you try to prove it yourself first before looking up a proof. Also, note that if you're dealing with two functions $\phi,\psi$, then by considering their difference $f=\phi-\psi$, you can obtain analogous theorems which tell you when $\int_a^b\phi\geq \int_a^b\psi$ and when $\int_a^b\phi>\int_a^b\psi$.
For your second doubt, I can't see immediately why $c\neq 0$ a-priori, because the mean-value theorem for integrals merely gives us some $c$, and it doesn't specify which one. So, I think your first approach is the way to go (and it's also the simplest); the only thing of course is you need to know the theorem I mentioned above (the proof isn't too hard anyway, and should be immediate from the definition of Riemann integrals).
A: Contemplate that in $[\frac13,\frac23]$ you strictly have
$$e^{-1}<e^{-x^2}<1$$ and by integration
$$\int_{1/3}^{2/3}e^{-1}\,dx<\int_{1/3}^{2/3}e^{-x^2}\,dx<\int_{1/3}^{2/3}\,dx.$$
Hence even if the functions are equal in the rest of the domain, the total contributions cannot be equal. This will work if inequality holds in an arbitrary subinterval, however tiny.
