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

Let $I =\int_1^2 \frac{\sqrt{x}}{\sqrt{3-x}+\sqrt{x}} d x$...(1)
$=\int_1^2 \frac{\sqrt{3-x}}{\sqrt{3-(3-x)}+\sqrt{3-x}} d x$
$\ldots \ldots\left\{\int_a^b f(x) d x=\int_a^b f(a+b-x) d x\right\}$
$I=\int_1^2 \frac{\sqrt{3-x}}{\sqrt{x}+\sqrt{3-x}} d x$...(2)
Adding (1) and (2)
$2 I =\int_1^2 \frac{\sqrt{x}+\sqrt{3-x}}{\sqrt{x}+\sqrt{3-x}} d x$
$=\int_1^2 1 d x=[x]_1^2$
$\therefore \quad 2 I =2-1=1$
$\therefore \quad I=\frac{1}{2}$

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