An agent's expected utility depends only on mean and variance
In order to understand this problem, I will work through the generic case. Say that a user had generalized quadratic (Bernoulli) utility, similar to your problem:

$$u(x) = \beta x^2 + \gamma x$$

and suppose that there is a distribution for the outcome of $x$, denoted $F(x)$. Thus, utility over this distribution is equal to

$$\begin{align}
\int u(x) \text{d}F(x) &amp; = \int (\beta x^2 + \gamma x) \text{d}F(x) \\
&amp; = \beta \int x^2 \text{d}F(x) + \gamma \int x \text{d}F(x) \\
&amp; = \beta \int x^2 \text{d}F(x) + \left[ - \beta \left(\int x \text{d}F(x)\right)^2 + \beta \left(\int x \text{d}F(x)\right)^2 \right] + \gamma \int x \text{d}F(x) \\
&amp; = \left[\beta \int x^2 \text{d}F(x) - \beta \left(\int x \text{d}F(x)\right)^2 \right] + \beta \left(\int x \text{d}F(x)\right)^2  + \gamma \int x \text{d}F(x) \\
&amp; = \beta\cdot(\text{variance of F(x)}) + \beta\cdot(\text{mean of F(x)})^2 + \gamma \cdot (\text{mean of F(x)})
\end{align}$$

So utility is determined by the mean and variance of the payoff distribution. If you understand the above work, then it should be easy to do the specific case you have given us. Try it for yourself.



Considering your specific case:

$$U(Y_s) = -\frac{1}{2}(\alpha - Y_s)^2$$

implies

$$U(L) = \sum_{s=1}^{S}\pi_s (-\frac{1}{2}(\alpha - Y_s)^2)$$

You can technically work with either finding $\int U(L) \text{d}L$ or $\int U(Y_s) \text{d}L$. Best of luck with your work.

Edit: Working in the discrete case will mean using expected values may prove helpful.