MCQ
$I = \int_0^{\pi /2} {\frac{{{{(\sin x + \cos x)}^2}}}{{\sqrt {1 + \sin 2x} }}{\rm{ }}} dx =$
- A$3$
- B$1$
- ✓$2$
- D$0$
$ = \int_0^{\pi /2} {\frac{{{{(\sin x + \cos x)}^2}}}{{\sqrt {{{(\sin x + \cos x)}^2}} }}dx} $
$I = \int_0^{\pi /2} {(\sin x + \cos x)dx = ( - \cos x + \sin x)_0^{\pi /2}} $
$I = 1 - ( - 1) = 2$.
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$I=\int_{0}^{10} \frac{[x] e^{[x]}}{e^{x-1}} d x,$
જ્યાં $[ x ]$ એ $x$ અથવા $x$ થી નાનો મહત્તમ પૂર્ણાંક દર્શાવે છે. તો $I$ $= .....$