The value of int_2^3 {frac{{sqrt x }}{{sqrt { | vedclass

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(D) Let $I = \int_2^3 {\frac{{\sqrt x }}{{\sqrt {5 - x} + \sqrt x }}} \,dx$ .....$(i)$Using the property $\int_a^b f(x) \,dx = \int_a^b f(a + b - x) \

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$\frac{1}{2}$

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(D) Let $I = \int_2^3 {\frac{{\sqrt x }}{{\sqrt {5 - x} + \sqrt x }}} \,dx$ .....$(i)$Using the property $\int_a^b f(x) \,dx = \int_a^b f(a + b - x) \,dx$,we replace $x$ with $(2 + 3 - x) = (5 - x)$.$	herefore I = \int_2^3 {\frac{{\sqrt {5 - x} }}{{\sqrt {5 - (5 - x)} + \sqrt {5 - x} }}} \,dx$$I = \int_2^3 {\frac{{\sqrt {5 - x} }}{{\sqrt x + \sqrt {5 - x} }}} \,dx$ .....$(ii)$Adding $(i)$ and $(ii)$:$2I = \int_2^3 {\frac{{\sqrt x + \sqrt {5 - x} }}{{\sqrt {5 - x} + \sqrt x }}} \,dx$$2I = \int_2^3 1 \,dx$$2I = [x]_2^3 = 3 - 2 = 1$$I = \frac{1}{2}$.

The value of $\int_2^3 {\frac{{\sqrt x }}{{\sqrt {5 - x} + \sqrt x }}} \,dx$ is

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