Integrate this function: $⌊x^2⌋ + \{x^2\}$ , from $0$ to $2$
I know that the first part is a step function, and every integer value, say, from 1 to 2, will be 1. But I can't understand how to integrate it. Also does the square matter if its outside the bracket or inside?
As Iulu stated in the comments, it does matter whether the square is on the outside or the inside, but as you've written the problem here, since $\lfloor z \rfloor +\{z\}=z$, we have $\lfloor x^2 \rfloor +\{x^2\}=x^2$. And thus we have $$\int_0^2 \lfloor x^2 \rfloor +\{x^2\} \ dx = \int_0^2 x^2=\frac{x^3}{3}\bigg|_0^2 = \frac{8}{3}$$
What may be slightly more interesting, however, would be an evaluation of the two parts individually. For $$\int_0^2 \lfloor x^2 \rfloor \ dx$$ we can split the integral into parts as follows. For $0<x<1$, we have $\lfloor x^2 \rfloor = 0$. For $1<x<\sqrt{2}$, we have $\lfloor x^2 \rfloor=1$, and in general we have for $\sqrt{n}<x<\sqrt{n+1}$, we have $\lfloor x^2 \rfloor =n$. So splitting up the integral we get $$\int_0^2 \lfloor x^2 \rfloor \ dx = \sum_{n=0}^3 \int_{\sqrt{n}}^{\sqrt{n+1}} n \ dx = \sum_{n=1}^{3}n(\sqrt{n+1}-\sqrt{n})=3\sqrt{4}-\sqrt{3}-\sqrt{2}-\sqrt{1}=5-\sqrt{3}-\sqrt{2}$$ Note that we can do a similar splitting of the integral for the fractional part, but I'll leave that as an optional excercise.