TITLE: How do I prove the inequality
QUESTION [5 upvotes]: $$1 - \frac{1}{n} < \int \limits_0^1 e^{-x^n}dx < 1$$
I probably have to apply the mean value theorem, but I don't know how to choose $f(x)$ and $g(x)$.
The other option is to do a partial integration, but $x$ here is in $n$th power.

REPLY [5 votes]: One has $1-y < e^{-y}$ for $y \neq 0$ by various arguments (perhaps the easiest is that $e^y$ is convex, so lies above its tangent, but one could equally employ the Mean Value Theorem to reach the same conclusion). Equally, $e^{-y}$ is decreasing, so for $y>0$ one has $e^{-y}<1$. So one has
$$ 1-x^{n} < e^{-x^n} < 1 \quad (x>0) $$
Integrating from $0$ to $1$ gives
$$ 1-\frac{1}{n+1} < \int_0^1 e^{-x^n} \, dx < 1, $$
which is stronger than the required inequality (the single point of equality at an endpoint clearly makes no difference, and $n<n+1 \implies 1-1/n<1-1/(n+1)$).