Evaluate the following integrals : _1^2 x 3-x+x d x - Vidyadip

Question

Evaluate the following integrals : $\int_1^2 \frac{\sqrt{x}}{\sqrt{3-x}+\sqrt{x}} d x$

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Answer

$
\begin{aligned}
& \text { Let } I =\int_1^2 \frac{\sqrt{x}}{\sqrt{3-x}+\sqrt{x}} \cdot d x \\
& =\int_1^2 \frac{\sqrt{1+2-x}}{\sqrt{3-(1+2-x)}+\sqrt{1+2-x}} \cdot d x \quad \ldots\left[\because \int_{ a }^{ b } f(x) \cdot d x=\int_{ a }^{ b } f( a + b -x) \cdot d x\right] \\
& \therefore I =\int_1^2 \frac{\sqrt{3-x}}{\sqrt{x}+\sqrt{3-x}} \cdot d x
\end{aligned}
$
Adding (i) and (ii), we get
$
\begin{aligned}
& 2 I =\int_1^2 \frac{\sqrt{x}}{\sqrt{3-x}+\sqrt{x}} \cdot d x+\int_1^2 \frac{\sqrt{3-x}}{\sqrt{x}+\sqrt{3-x}} \cdot d x \\
& =\int_1^2 \frac{\sqrt{x}+\sqrt{3-x}}{\sqrt{x}+\sqrt{3-x}} \cdot d x \\
& =\int_1^2 1 \cdot d x \\
& =[x]_1^2 \\
& \therefore 2 I =2-1=1 \\
& \therefore I =\frac{1}{2} .
\end{aligned}
$

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