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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Find $ \int_2^8 \large\frac{\sqrt{10-x}}{\sqrt x+\sqrt{10-x}}$$dx $

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Toolbox:
  • $\int\limits_a^bf(x)dx=\int\limits_a^b[ f(a+b)-x]dx$
Step 1:
$I=\int\limits_2^8\large\frac{\sqrt{10-x}}{\sqrt x+\sqrt{10-x}}$$dx$------(1)
$\int\limits_a^bf(x)dx=\int\limits_a^b[ f(a+b)-x]dx$
Applying the above property we get,
$I=\int\limits_2^8\large\frac{\sqrt{10-[(8+2)-x]}}{\sqrt{8+2-x}+\sqrt{10-(8+2-x)}}$
$\;\;=\int\limits_2^8\large\frac{\sqrt{10-10+x}}{\sqrt{10-x}+\sqrt{10-10+x}}$$dx$
$I=\int\limits_2^8\large\frac{\sqrt x}{\sqrt{10-x}+\sqrt x}$$dx$-------(2)
Step 2:
Add equation(1) and equation(2)
$2I=\int\limits_2^8\large\frac{\sqrt x+\sqrt{10-x}}{\sqrt{10-x}+\sqrt x}$
$\;\;\;\;=\int\limits_2^8dx$
On integrating we get,
$\therefore 2I=\big[x\big]_2^8$
Step 3:
On applying limits we get,
$2I=[8-2]$
$2I=6$
$I=3$
answered Nov 7, 2013 by sreemathi.v
 
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