Question Bank [2022] — Maths STD 12 Science — Question
Maharashtra BoardEnglish MediumSTD 12 ScienceMathsQuestion Bank [2022]2 Marks
Question
Evaluate: $\int_0^9 \frac{\sqrt{x}}{\sqrt{x}+\sqrt{9-x} d x}$
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Answer
$ \text { Let } I =\int_0^9 \frac{\sqrt{x}}{\sqrt{x}+\sqrt{9}-x} d x \quad \ldots \ldots . . \text { (i) }$
$=\int_0^9 \frac{\sqrt{9-x}}{\sqrt{9-x}+\sqrt{9-(9-x)}} d x \quad \ldots \ldots\left[\because \int_0^{ a } f (x) d x=\int_0^{ a } f ( a -x) d x\right]$
$\therefore I =\int_0^9 \frac{\sqrt{9-x}}{\sqrt{9-x}+\sqrt{x}} d x \quad \ldots \ldots \text { (ii) } $
Adding (i) and (ii), we get
$ 2 l =\int_0^9 \frac{\sqrt{x}}{\sqrt{x}+\sqrt{9-x}} d x+\int_0^9 \frac{\sqrt{9-x}}{\sqrt{9-x}+\sqrt{x}} d x$
$=\int_0^9 \frac{\sqrt{x}+\sqrt{9-x}}{\sqrt{x}+\sqrt{9-x}} d x$
$=\int_0^9 d x$
$=[x]_0^9$
$\therefore 2 l =9-0$
$\therefore I =\frac{9}{2} $
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