Expectation of Minimum of $n$ i.i.d. uniform random variables.

To calculate the expected value, we're going to need the density function for $Y$. To get that, we're going to need the distribution function for $Y$. Let's start there.

By definition, $F(y) = P(Y \leq y) = 1 - P(Y > y) = 1 - P(\text{min}(X_1, \ldots, X_n) > y)$. Of course, $\text{min}(X_1, \ldots X_n) > y$ exactly when $X_i > y$ for all $i$. Since these variables are i.i.d., we have $F(y) = 1 - P(X_1 > y)P(X_2>y)\ldots P(X_n>y) = 1 - P(X_1 > y)^n$. Assuming the $X_i$ are uniformly distributed on $(a, b)$, this yields $$F(y) = \left\{ \begin{array}{ll} 1 - \left(\frac{b-y}{b-a}\right)^n & : y \in (a, b)\\ 0 & : y < a\\ 1 & : y > b \end{array} \right.$$ We take the derivative to get the density function. $$f(y) = \left\{ \begin{array}{ll} \frac{n}{b-a} \left(\frac{b-y}{b-a}\right)^{n-1} & : y \in (a, b)\\ 0 & : \text{otherwise} \end{array} \right.$$ Now $E(Y) = \int_{-\infty}^{\infty} y f(y) dy$. The integral is straightforward; I'll leave the details to you. I calculate $E(Y) = \frac{b+na}{n+1}$.

Yes. Assuming a $U(0,1)$, note that

$$\Pr\Bigl(\min_i X_i\leq x\Bigr) = 1- \Pr\Bigl(\min_i X_i\geq x\Bigr) = 1- (1-x)^n. $$

So the density function is

$$f(x)=n(1-x)^{n-1}.$$

Then

$$\int_0^1 x f(x) dx = n\int_0^1 x(1-x)^{n-1} dx = n\int_0^1 (1-t) t^{n-1} dt = \frac{1}{n+1}. $$

An short proof without the use of the pdf. Suppose $X_i \sim U(0,1)$. Then \begin{align*} E[\min(X_i)] &= \int_0^1 P[\min(X_i)>\mu] d\mu \\ &= \int_0^1 \prod_i P[X_i>\mu] d\mu \\ &= \int_0^1 (1-\mu)^n d\mu\\ &= \int_0^1 \mu^n d\mu\\ &= \dfrac{1}{n+1} \end{align*}