MCQ
$\int_{\,0}^{\,\sqrt 2 } {[{x^2}]\,dx} =$ (કે જ્યાં $[.]=$ એ મહતમ પૂર્ણાક છે )
- A$2 - \sqrt 2 $
- B$2 + \sqrt 2 $
- ✓$\sqrt 2 - 1$
- D$\sqrt 2 - 2$
$ = \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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