$I=\int_{-1}^{2}\frac{xf(x^2)}{2+f(x+1)}dx$

Question

Let $f:R\to R$ be a function defined by $f(x)=\begin{cases} [x] & x\leq 2 \\ 0 & x>2 \\ \end{cases}$
where $[x]$ is the greatest integer less than or equal to $x$.Then find $I=\int_{-1}^{2}\frac{xf(x^2)}{2+f(x+1)}dx$

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This question was asked in JEE Advanced.I have problem in breaking the limits of integration into parts.
I broke the integration limits like this.
$I=\int_{-1}^{2}\frac{xf(x^2)}{2+f(x+1)}dx=\int_{-1}^{0}\frac{x\bigl[x^2\bigr]}{2+\bigl[x+1\bigr]}dx+\int_{0}^{1}\frac{x\bigl[x^2\bigr]}{2+\bigl[x+1\bigr]}dx+\int_{1}^{\sqrt{2}}\frac{x\bigl[x^2\bigr]}{2+\bigl[x+1\bigr]}dx+\int_{\sqrt{2}}^{\sqrt{3}}\frac{x\bigl[x^2\bigr]}{2+\bigl[x+1\bigr]}dx+\int_{\sqrt{3}}^{2}\frac{x\bigl[x^2\bigr]}{2+\bigl[x+1\bigr]}dx$

But the book says
$I=\int_{-1}^{2}\frac{xf(x^2)}{2+f(x+1)}dx=\int_{-1}^{0}\frac{x\bigl[x^2\bigr]}{2+\bigl[x+1\bigr]}dx+\int_{0}^{1}\frac{x\bigl[x^2\bigr]}{2+\bigl[x+1\bigr]}dx+\int_{1}^{\sqrt{2}}\frac{x\bigl[x^2\bigr]}{2+\bigl[x+1\bigr]}dx$

I am confused why have they not included limits from $\sqrt{2}$ to $\sqrt{3}$ and limits from $\sqrt{3}$ to $2.$After splitting of limits,i could solve.Please help me.

Answer 1

Above $x=\sqrt2$, $f(x^2)$ doesn't bring any contribution as it is zero.