Question
$\int_{\,0}^{\,\sqrt 2 } {[{x^2}]\,dx} ,$ (जहाँ $[.]$ एक महतम पूर्णाक फलन है)
$ = \int_{\,0}^{\,1} {[{x^2}]\,dx + } \int_{\,1}^{\,\sqrt 2 } {[{x^2}]\,\,dx} $
$ = \int_{\,0}^{\,1} {\,0\,dx + } \int_{\,1}^{\,\sqrt 2 } {\,dx} $
$ = [x]_1^{\sqrt 2 } = \sqrt 2 - 1$.
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