Evaluate _1^2 x 3-x+x d x - Vidyadip

Question

Evaluate $\int_1^2 \frac{\sqrt{x}}{\sqrt{3-x}+\sqrt{x}} d x$

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Answer

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

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