Expected value of the maximum of 3 independent random variables

$\\\text{Three individuals are running a one kilometer race. The completion time for each individual is a random variable. }X_i \text{ is the completion time, in minutes, for person }i.\\\\X_1 \text{: uniform distribution on the interval }[2.9,3.1]\\\\X_2\text{: uniform distribution on the interval }[2.7,3.1]\\\\X_3\text{: uniform distribution on the interval }[2.9,3.3]\\\\\text{The three completion times are independent of one another. Find the expected latest completion time.}$

I'm having trouble understanding the part of the solution that is between the asterisks.

$\\Y=max(X_1,X_2,X_3).\text{ }f_Y(y)=F'_Y(y)\text{, where}\\\\F_Y(y)=P[Y\leq y]=P[max(X_1,X_2,X_3)\leq y]\\\\=P[(X_1\leq y)\cap(X_2\leq y)\cap(X_3\leq y)]\\\\=P[X_1\leq y]*P[X_2\leq y]*P[X_3\leq y]\\\\$

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$\\\begin{cases}5(y-2.9)*2.5(y-2.7)*2.5(y-2.9)=31.25(y^3-8.5y^2+24.07y-22.707)& \text{ for }2.9\leq y\leq 3.1\\2.5(y-2.9)& \text{ for }3.1\leq y\leq 3.3\end{cases}$

$\\\text{and }F_Y(y)=0\text{ for } y\leq 2.9.$
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$\\\text{Then, } f_Y(y)=F'_Y(y)=\begin{cases}31.25(3y^2-17y+24.07)& \text{ for }2.9\leq y\leq 3.1\\2.5& \text{ for }3.1 \leq y \leq 3.3\end{cases}$

$\\\text{Finally, }E[Y]=\int^{3.1}_{2.9}y*31.25(3y^2-17y+24.07)dy+\int^{3.3}_{3.1}y*2.5dy=\frac{73}{48} +1.6=3.12$

I think my biggest question is, why is the integral split up instead of being over [2.7,3.3]?