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Evaluate:\[\int_{-1}^1 log \bigg(\large\frac{4-x}{4+x}\bigg)dx\]

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  • (i)$\int \limits_a^b f(x)dx=F(b)-F(a)$
  • (ii)$ \int _{-a}^a f(x) dx =0 $ if f(x) is an odd function
  • (iii)If $f(-x)= -f(x),$ then it is an odd function
Given $I= \int \limits_{-1}^1 log \frac{(4-x)}{(4+x)}dx$
Let $ f(x)=log \frac{(4-x)}{(4+x)}$ then $ f(x)=log \bigg(\frac{4+x}{4-x}\bigg)$
If x is replaced by -x
then $f(-x)=log \bigg(\frac{4+x}{4-x} \bigg)=-f(x)$
Hence it is an odd function
$ \int \limits_{-a}^a f(x) =0$
Therefore $ I=0$



answered Mar 15, 2013 by meena.p
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