Question
Evaluate the following integrals:
$\int\limits^{\sqrt{2}}_0\big[\text{x}^2\big]\text{dx}$

Answer

We have,
$\text{I}=\int\limits^{\sqrt{2}}_0\big[\text{x}^2\big]\text{dx}$
$=\int\limits^{1}_0\big[\text{x}^2\big]\text{dx}+\int\limits^{\sqrt{2}}_1\big[\text{x}^2\big]\text{dx}$
$=\int\limits^{1}_0(0)\text{dx}+\int\limits^{\sqrt{2}}_1(1)\text{dx}$ $\begin{pmatrix}\because\big[\text{x}\big]^2=\begin{cases}0,&0<\text{x}<1\\1,&1<\text{x}<\sqrt{2}\end{cases}\end{pmatrix}$ 
$=0+\big[\text{x}\big]^{\sqrt{2}}_1$
$=\sqrt{2}-1$

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