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$\int_3^6 \frac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}} d x=$
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Verified Answer
The correct answer is:
$\frac{3}{2}$
$I=\int_3^6 \frac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}} d x$ ...(i)
$\because \int_a^b f(x) d x=\int_a^b f(a+b-x) d x$
$I=\int_3^6 \frac{\sqrt{9-x}}{\sqrt{x}+\sqrt{9-x}} d x$ ...(ii)
Add (i) and (ii),
$\begin{aligned} & 2 I=\int_3^6 1 d x=[x]_3^6 \\ & \Rightarrow 2 I=3 \Rightarrow I=\frac{3}{2}\end{aligned}$
$\because \int_a^b f(x) d x=\int_a^b f(a+b-x) d x$
$I=\int_3^6 \frac{\sqrt{9-x}}{\sqrt{x}+\sqrt{9-x}} d x$ ...(ii)
Add (i) and (ii),
$\begin{aligned} & 2 I=\int_3^6 1 d x=[x]_3^6 \\ & \Rightarrow 2 I=3 \Rightarrow I=\frac{3}{2}\end{aligned}$
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