Mathematics

$$\int_{0}^{\sqrt{2}} [x^{2}] dx$$ is equal to (where $$[.]$$ denotes greatest integer function )


ANSWER

$$\sqrt {2}-1$$


SOLUTION
$$f(x)=\begin{cases} 0\,\,\,\,\,\,\,\,0\le x<1 \\1\,\,\,\,\,\,\,\le x<\sqrt{2}  \end{cases}$$
$$\displaystyle \int^{\sqrt{2}}_0 [x^2]dx=\int^1_0 0.dx+\int^{\sqrt{2}}_1 1.dx$$
                     $$=0+[x]^{\sqrt{2}}_1$$
                     $$=\sqrt{2}-1$$
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Single Correct Medium Published on 17th 09, 2020
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