How can $\int_{-3}^3\lfloor{x^2}\rfloor dx = 2 \cdot \int_0^3\lfloor{x^2}\rfloor dx$? This website simplifies the integral $$\int_{-3}^3\lfloor{x^2}\rfloor dx$$
into
$$2 \cdot \int_0^3\lfloor x^2 \rfloor dx.$$
How is that done? I'm unsure which property of the integral it is.
The closest I have found is the Homogeneous Property, which states $$\int_a^b c \cdot s(x) dx = c \cdot \int_a^b s(x) dx.$$
Also, note that we are dealing with step functions, and by $\lfloor x^2 \rfloor$, I mean the greatest integer $\le x^2$.
Any insight you could provide as to of how that simplification is made would be much appreciated. Thank you.
 A: Notice first that if we define some step function $s(x) = \lfloor x^2 \rfloor$, then $s(x)$ is even, since for all $x$ in the domain of $s$, $s(x) = s(-x)$.
We shall use the property Additivity with Respect to the Interval of Integration to help deduce our solution, which states
$$\int_a^c f(x) dx + \int_c^b f(x) dx = \int_a^b f(x)dx$$
where $a < c < b$.
In this case, taking $a$, $c$, and $b$ to be $-3$, $0$, and $3$ respectively, we may say
$$\int_{-3}^3 \lfloor x^2 \rfloor dx = \int_{-3}^0 \lfloor x^2 \rfloor dx + \int_0^3 \lfloor x^2 \rfloor dx.$$
Observe, then, that since $s(x)$ is even, it's graph is symmetric on both sides of the $y$-axis. Due to this congruency, 
$$\int_{-3}^0 \lfloor x^2 \rfloor dx = \int_0^3 \lfloor x^2 \rfloor dx.$$
That means, then, that we are adding the same value twice to compute the full integral. By simple algebra, we may rewrite 
$$\int_{-3}^3 \lfloor x^2 \rfloor dx $$
as 
$$2 \cdot \int_0^3 \lfloor x^2 \rfloor$$
More generally, when ever we are integrating an even function $f(x)$ along the closed interval $[-a, a]$, we may say
$$\int_{-a}^a f(x) dx = 2 \cdot \int_0^af(x)dx.$$
