Question

__________________??

Answer 1

$\textcolor{red}{SOLUTION}$

$let \: I \: = \displaystyle \int\limits_{0}^{6} { \frac{ \sqrt{x} }{ \sqrt{6 - x} + \sqrt{x} } } \, dx - - (1)$

$we \: know \: that \: \displaystyle \int\limits_{0}^{a} {f(x)} \, dx = \int\limits_{0}^{a} {f(a - x)} \, dx$

HENCE

$I = \displaystyle \int\limits_{0}^{6} { \frac{ \sqrt{6 - x} }{ \sqrt{6 - x} + \sqrt{x} } } \, dx - - - - - - (2)$

Now Equation (1) + Equation (2) gives

$I + I \: = \displaystyle \int\limits_{0}^{6} { \frac{ \sqrt{x} }{ \sqrt{6 - x} + \sqrt{x} } } \, dx + \displaystyle \int\limits_{0}^{6} { \frac{ \sqrt{6 - x} }{ \sqrt{6 - x} + \sqrt{x} } } \, dx$

$2I \: = \displaystyle \int\limits_{0}^{6} { \frac{ \sqrt{x} + \sqrt{6 - x} }{ \sqrt{6 - x} + \sqrt{x} } } \, dx$

$2I \: = \displaystyle \int\limits_{0}^{6} { 1 } \, dx$

$2I = \left. x\right]^6_0$

$2I = 6$

$I = 3$

HENCE

$\displaystyle \int\limits_{0}^{6} { \frac{ \sqrt{x} }{ \sqrt{6 - x} + \sqrt{x} } } \, dx \: = 3$

Answer 2

Refer from the attachment provided... !!!

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