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By using the properties of definite integrals,evaluate the integral\[\int\limits_0^a\frac{\sqrt x}{\sqrt x+\sqrt{a-x}}dx\]

$\begin{array}{1 1}a \log 2 \\\frac{a}{2} \\ \frac{a}{2}\log 2 \\ \large\frac{a}{4} \end{array} $

1 Answer

Toolbox:
  • (i)$\int\limits_a^b f(x)dx=F(b)-F(a)$
  • $\int\limits_0^a f(x)dx=\int \limits_0^a f(a-x)dx$
Given $I=\int\limits_0^a\frac{\sqrt x}{\sqrt x+\sqrt{a-x}}dx-----(1)$
 
Now applying the property $\int\limits_0^a f(x)dx=\int \limits_0^a f(a-x)dx$
 
Given $I=\int\limits_0^a\frac{\sqrt {a-x}}{\sqrt {a-x}+\sqrt{a-a+x}}dx-----(2)$
 
Now adding equ(1) and equ(2)
 
Given $2I=\int\limits_0^a\frac{\sqrt x+\sqrt{a-x}}{\sqrt x+\sqrt{a-x}}dx=\int \limits_0^a dx$
 
$2I=\int \limits_0^adx$
 
On integrating we get
 
$2I=[x]_0^a$
 
On applying the limits,
 
$2I=a-0$
 
$=a$
 
Therefore $I=\frac{a}{2}$

 

answered Feb 14, 2013 by meena.p
 

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