Definite integral of ceiling function

Question

I understand the basics of solving definite integrals but I have trouble with the following one because of the ceiling function. How can I solve this?

$\int_0^3 [x^2] dx$

Answer 1

We have

$$x^2-1<\lfloor x^2\rfloor=n\le x^2\iff \sqrt n\le x<\sqrt{n+1}$$ so for $n=0$ we have $0\le x< 1$ and $\lfloor x^2\rfloor=0$ and for $n=1$ we have $1\le x<\sqrt2$ and $\lfloor x^2\rfloor=1$ and so on so we get

$$\int_0^3 \lfloor x^2\rfloor dx=\int_0^1\lfloor x^2\rfloor dx+\int_1^{\sqrt 2}\lfloor x^2\rfloor dx+\int_{\sqrt 2}^{\sqrt3}\lfloor x^2\rfloor dx+\cdots+\int_{\sqrt 8}^{3}\lfloor x^2\rfloor dx\\=\int_0^10dx+\int_1^{\sqrt 2}1 dx+\int_{\sqrt 2}^{\sqrt3}2 dx+\cdots+\int_{\sqrt 8}^{3}8 dx$$ Can you take it from here?

Answer 2

Hint: Over the interval $[0,3)$, $\lfloor x^2\rfloor \in[0,8]$. Given $k\in[0,8]$, what is the length of the interval in which $\lfloor x^2\rfloor = k$?