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By using the properties of definite integrals, evaluate the integrals
$\int_0^a \frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}} \mathrm{dx}$
$\int_0^a \frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}} \mathrm{dx}$
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Verified Answer
Let $\mathrm{I}=\int_0^a \frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}} d x \quad \ldots$ (i)
$\begin{aligned}
&\text { Then, } \mathrm{I}=\int_0^a \frac{\sqrt{a-x}}{\sqrt{a-x}+\sqrt{a-(a-x)}} d x \\
&\Rightarrow \mathrm{I}=\int_0^a \frac{\sqrt{a-x}}{\sqrt{a-x}+\sqrt{x}} d x \quad \ldots (ii)
\end{aligned}$
Adding (i) and (ii), we get ;
$2 \mathrm{I}=\int_0^a 1 d x=[x]_0^a=a-0=a \quad \therefore \mathrm{I}=\frac{a}{2}$
$\begin{aligned}
&\text { Then, } \mathrm{I}=\int_0^a \frac{\sqrt{a-x}}{\sqrt{a-x}+\sqrt{a-(a-x)}} d x \\
&\Rightarrow \mathrm{I}=\int_0^a \frac{\sqrt{a-x}}{\sqrt{a-x}+\sqrt{x}} d x \quad \ldots (ii)
\end{aligned}$
Adding (i) and (ii), we get ;
$2 \mathrm{I}=\int_0^a 1 d x=[x]_0^a=a-0=a \quad \therefore \mathrm{I}=\frac{a}{2}$
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