Question
By using the properties of definite integrals, evaluate the integral $\int\limits_0^a {\frac{{\sqrt x }}{{\sqrt x + \sqrt {a - x} }}} dx$

Answer

Let $I = \int\limits_0^a {\frac{{\sqrt x }}{{\sqrt x + \sqrt {a - x} }}} dx$…(i)
$\Rightarrow I = \int\limits_0^a {\frac{{\sqrt {a - x} }}{{\sqrt {a - x} + \sqrt {a - \left( {a - x} \right)} }}dx} $
$\left[ {\because \int\limits_0^a {f\left( x \right)dx = \int\limits_0^a {f\left( {a - x} \right)dx = } } } \right]$
$= \int\limits_0^a {\frac{{\sqrt {a - x} }}{{\sqrt {a - x} + \sqrt x }}} dx$ …(ii)
Adding eq. (i) and (ii),

$2I = \int\limits_0^a {\left( {\frac{{\sqrt x }}{{\sqrt x + \sqrt {a - x} }} + \frac{{\sqrt {a - x} }}{{\sqrt {a - x} + \sqrt x }}} \right)dx} $
$= \int\limits_0^a {\left( {\frac{{\sqrt x + \sqrt {a - x} }}{{\sqrt x + \sqrt {a - x} }}} \right)dx}$
$= \int\limits_0^a {1dx} = \left( x \right)_0^a = a$
$ \Rightarrow I = \frac{a}{2}$

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