Expected value of a minimum 
Under a group insurance policy, an insurer agrees to pay 100% of the medical bills incurred during the year by employees of a company, up to a max of $1$ million dollars. The total bills incurred, $X$, has pdf
  $$
f_X(x) =
\begin{cases}
\frac{x(4-x)}{9}, & \text{for } 0 < x < 3\\
0& \text{else}
\end{cases}
$$
  where $x$ is measured in millions. Calculate the total amount, in millions of dollars, the insurer would expect to pay under this policy.

So I was able to obtain part of the solution, which was
$$
E(\min(X,1)) = \int_0^1 x\cdot \frac{x(4-x)}{9} dx \tag1
$$
However, the solution has $(1)$, plus 
$$
E(\min(X,1)) = \int_1^3 1\cdot \frac{x(4-x)}{9} dx \tag2
$$
What I don't understand is if the problem explicitly states that they agree to pay up to $1$ million, why would you even have to bother with $(2)$?

For amounts greater than one million, they still pay out a million.
Let $x$ be the total bills in millions. For $x \in [0,1]$ the payout is $x$, for $x \in [1,\infty)$, the payout is $1$. Hence the payout as a function of $x$ is $p(x) = \min(x,1)$, and you wish to compute $Ep$.
\begin{eqnarray}
Ep &=& \int_0^\infty p(x) f_X(x) dx \\
&=& \int_0^1 x f_X(x) dx + \int_1^3 1 f_X(x)dx  + \int_3^\infty 1 f_X(x)dx \\
&=& \frac{13}{108} + \frac{22}{27}+ 0 \\
&=& \frac{101}{108}
\end{eqnarray}
