Let $U : \mathbb{R} \times \mathbb{R} \nrightarrow \mathrm{dist}[\mathbb{R}]$ denote the parametrized family of uniform distributions where $U(a, b)$ is the uniform distribution with minimum $a$ and maximum $b$ . $U$ is a partial function defined whenever $a \lt b$ .
Let's define $Y_1, Y_2, \dots Y_n$ as an $n$-element sample drawn from $U(0,1)$ . Let $Y_{(1)}$ denote the sample minimum. Let $f$ denote the pdf of the sample minimum of $n$ elements drawn from $U(0, 1)$ and let $F$ denote the corresponding cdf.
I want to demonstrate that its expected value is $\frac{1}{n+1}$ in a way that's as simple as possible, ideally without using calculus.
$$ \mathrm{E}[Y_{(1)}] = \frac{1}{n+1} $$
Here's one way to do it, which does use calculus:
$$ \mathrm{E}[Y_{(1)}] \tag{1}$$
use known formula for expectation in terms of cdf
$$ \int_{s=0}^\infty 1-F(s) \mathrm{d}s \tag{2}$$
$F(s) = 1$ when $s \ge 1$ .
$$ \int_{s=0}^{1} 1 - F(s) \mathrm{d}s \tag{3} $$
$F(t)$ is the probability that the sample minimum is less than $t$.
$$ \int_{s=0}^{1} 1 - \mathbb{P}[Y_1 \le s \lor \cdots \lor Y_n \le s] \mathrm{d}s \tag{4} $$
$1-\mathbb{P}[\psi]$ is $\mathbb{P}[\lnot \psi]$ .
$$ \int_{s=0}^{1} \mathbb{P}[Y_1 \ge s \land \cdots \land Y_n \ge s] \mathrm{d}s \tag{5} $$
replace integrand with product.
$$ \int_{s=0}^{1} (1-s)^n \mathrm{d}s \tag{6} $$
change variable $ t = 1-s $ and swap bounds of integral.
$$ \int_{t=0}^{1} t^n \mathrm{d}t \tag{7} $$
simplify
$$ \frac{1}{n+1} \tag{8} $$
ΟΕΔ
Take a circle of length $1$, and choose $n+1$ points independently and uniformly on it. Cut the circle at the first point, and unwrap to the interval $[0,1]$; the remaining $n$ points are distributed uniformly and independently on that interval. Now, those points cut $[0,1]$ into $n+1$ subintervals, in order. What's the expected length of each of them? Well, pull back to the circle - there was nothing special about us cutting at the first point. Cut at one of the other points, and we cycle the subintervals around, while keeping the same picture of $n$ uniform points after the cut. Thus, the lengths of all $n+1$ subintervals are identically distributed. In particular, they have the same mean. As their sum is $1$, the mean of each must be $\frac1{n+1}$ by linearity of expectation. Done.
This also works to calculate the expected value of all of the other order statistics; the $k$th smallest has expected value $\frac{k}{n+1}$.