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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Evaluate:$\int \limits_3^9 \frac {\sqrt {12-x}}{\sqrt x+\sqrt {12-x}}dx$

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Toolbox:
  • (i)$\int \limits _a^b f(x)dx=F(b)-F(a)$
  • (ii) $ \int \limits_a^b f(x)dx=\int f(a+b-x) dx$
Given $I=\int \limits_3^9 \large\frac {\sqrt {12-x}}{\sqrt x+\sqrt {12-x}}dx$-----(1)
By applying the property
$ \int \limits_a^b f(x)dx=\int f(a+b-x) dx$
$I=\int \limits_3^9 \large\frac {\sqrt {12-(9+3-x)}}{\sqrt {9+3-x+}+\sqrt {12-(9+3-x)}}dx=\large\frac{\sqrt x}{\sqrt {12-x}+\sqrt x}dx$
Therefore $I=\large\frac{\sqrt x}{\sqrt {12-x}+\sqrt x}$-----(2)
Adding equ 1 and equ 2
$2I=\int \limits_3^9 \large\frac {\sqrt {12-x}}{\sqrt x+\sqrt {12-x}}+\frac{\sqrt x}{\sqrt {12-x}+\sqrt x}dx$
$2I=\int \limits_3^9 \large\frac {\sqrt {12-x}+\sqrt x}{\sqrt {12-x}+\sqrt x}dx$
$2I=\int \limits_3^9 dx$
On integrating we get
$2I=\bigg[x\bigg]_3^9$
On applying limits we get
$2I=[9-3]=6$
$2I=6$
Therefore $I=\large\frac{6}{2}=3$

 

answered Apr 2, 2013 by meena.p
edited Apr 2, 2013 by meena.p
 
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