EXPLANATION.

[tex]\sf \displaystyle \int\limits^4_2 \frac{\sqrt{x} }{\sqrt{6 - x} + \sqrt{x} } dx[/tex]

As we know that,

King property.

[tex]\sf \displaystyle \int\limits^b_a f(x)dx = \int\limits^b_a f(a + b - x)dx[/tex]

Using this property in this question, we get.

[tex]\sf \displaystyle I = \int\limits^4_2 \frac{\sqrt{x} }{\sqrt{6 - x} + \sqrt{x} } dx. - - - - - (1)[/tex]

[tex]\sf \displaystyle I = \int\limits^4_2 \frac{\sqrt{(2 + 4 - x)} }{\sqrt{6 - (2 + 4 - x)} + \sqrt{(2 + 4 - x)} } dx[/tex]

[tex]\sf \displaystyle I = \int\limits^4_2 \frac{\sqrt{6 - x} }{\sqrt{6 - (6 - x)} + \sqrt{6 - x} } dx[/tex]

[tex]\sf \displaystyle I = \int\limits^4_2 \frac{\sqrt{6 - x} }{\sqrt{6 - 6 + x} + \sqrt{6 - x} } dx[/tex]

[tex]\sf \displaystyle I = \int\limits^4_2 \frac{\sqrt{6 - x} }{\sqrt{x} + \sqrt{6 - x} } dx. - - - - - (2)[/tex]

Adding expression (1) and (2), we get.

[tex]\sf \displaystyle 2I = \int\limits^4_2 \frac{\sqrt{x} + \sqrt{6 - x} }{\sqrt{6 - x} + \sqrt{x} } dx[/tex]

[tex]\sf \displaystyle 2I = \int\limits^4_21.dx[/tex]

[tex]\sf \displaystyle 2I = \bigg[ x \bigg]_{2}^{4}[/tex]

As we know that,

In definite integration first we put upper limits then we put lower limits in the expression, we get.

[tex]\sf \displaystyle 2I = (4 - 2)[/tex]

[tex]\sf \displaystyle 2I = 2[/tex]

[tex]\sf \displaystyle I = 1[/tex]

[tex]\sf \displaystyle \boxed{\int\limits^4_2 \frac{\sqrt{x} }{\sqrt{6 - x} + \sqrt{x} } dx = 1}[/tex]

MORE INFORMATION.

Queen property.

[tex]\sf \displaystyle \int_{0}^{2a} f(x)dx = \int\limits^a_0 f(x) + f(2a - x )dx[/tex]

[tex]\sf (a) = \displaystyle 2 \int\limits^a_0 f(x)dx \ \ \ if \ \ \ f(2a - x) = f(x)[/tex]

[tex]\sf (b) = \displaystyle 0 \ \ \ if \ \ \ f(2a - x) = - f(x)[/tex]

Verified answer

Hello dipyan!

Information provided:

[tex] \large\displaystyle {\sf{ \bull \int^4_2 \frac{ \sqrt{x} }{ \sqrt{6 - x} + \sqrt{x} } = dx}}[/tex]

Formula used:

Kings property-

[tex]\large \displaystyle{\bf{ \bull \int _{a}^{b} f(x)dx = \int_{a}^{b}f(a + b - x)dx }}[/tex]

Solution:

[tex] \large{\displaystyle { \implies\sf{\int^4_2 \frac{ \sqrt{x} }{ \sqrt{6 - x} + \sqrt{x} } dx}}}[/tex]

[tex] \large{\displaystyle { \implies I=\sf{\int^4_2 \frac{ \sqrt{x} }{ \sqrt{6 - x} + \sqrt{x} } dx - - - (1)}}}[/tex]

[tex] \large{\displaystyle { \implies I=\sf{\int^4_2 \frac{ \sqrt{2 + 4 - x} }{ \sqrt{6 - (2 + 4 - x)} + \sqrt{2 + 4 - x} } dx }}}[/tex]

[tex] \large{\displaystyle { \implies I=\sf{\int^4_2 \frac{ \sqrt{6- x} }{ \sqrt{6 - ( 6 - x)} + \sqrt{6- x} } dx }}}[/tex]

[tex] \large{\displaystyle { \implies I=\sf{\int^4_2 \frac{ \sqrt{6- x} }{ \sqrt{6 - 6 + x} + \sqrt{6- x} } dx }}}[/tex]

[tex] \large{\displaystyle { \implies I=\sf{\int^4_2 \frac{ \sqrt{6- x} }{ \sqrt{ x} + \sqrt{6- x} } dx - - - (2)}}}[/tex]

Add the expressions (1) and (2),

[tex] \large{\displaystyle { \implies I +I =\sf{\int^4_2 \frac{ \sqrt{x} + \sqrt{6- x} }{ \sqrt{6- x} + \sqrt{x} } dx}}}[/tex]

[tex] \large{\displaystyle { \implies 2I =\sf{\int^4_{2}1.dx}}}[/tex]

[tex] \large{\displaystyle { \implies 2I =\sf{ \bigg[x\bigg]_{2}^{4} }}}[/tex]

As it is a definite integration we will subtract the lower limit from upper limit.

[tex] \large{\displaystyle { \implies 2I =\sf{4 - 2 }}}[/tex]

[tex] \large{\displaystyle { \implies 2I =\sf{2}}}[/tex]

[tex] \large{\displaystyle { \implies I =\sf{ \frac{2}{2}}}}[/tex]

[tex] \large{\displaystyle { \implies I =\sf{1}}}[/tex]