int_3^6 frac{sqrt{x}}{sqrt{9-x}+sqrt{x}} d x= | vedclass

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(B) Let $I = \int_3^6 \frac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}} d x$ ...$(i)$Using the property $\int_a^b f(x) d x = \int_a^b f(a+b-x) d x$,where $a=3$ and

Vedclass · Accepted Answer

$\frac{3}{2}$

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(B) Let $I = \int_3^6 \frac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}} d x$ ...$(i)$Using the property $\int_a^b f(x) d x = \int_a^b f(a+b-x) d x$,where $a=3$ and $b=6$,we have $a+b-x = 9-x$.Substituting this into the integral:$I = \int_3^6 \frac{\sqrt{9-x}}{\sqrt{9-(9-x)}+\sqrt{9-x}} d x$$I = \int_3^6 \frac{\sqrt{9-x}}{\sqrt{x}+\sqrt{9-x}} d x$ ...(ii)Adding equations $(i)$ and (ii):$2I = \int_3^6 \frac{\sqrt{x}+\sqrt{9-x}}{\sqrt{x}+\sqrt{9-x}} d x$$2I = \int_3^6 1 d x$$2I = [x]_3^6 = 6 - 3 = 3$$I = \frac{3}{2}$

$\int_3^6 \frac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}} d x=$

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