Mathematics

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

$\sqrt {2}-1$

##### 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
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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