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Home  >>  CBSE XII  >>  Math  >>  Integrals

Evaluate the definite integral\[\int\limits_0^1\frac{dx}{1+x^2}\]

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Toolbox:
  • $\int \limits_a^b f(x)dx=F(b)-F(a)$
  • $\int\large\frac{dx}{1+x^2} dx=\tan^{-1}(x)+c$
Given $\int\limits_0^1\Large\frac{dx}{1+x^2}$
 
On integrating we get,
]
$I=\bigg[\tan ^{-1}(x)\bigg]^1_0$
 
On Applying limits we get.
 
$[\tan ^{-1}(1)-\tan ^{-1}(0)]$
 
But $ \tan^{-1}(1)=\frac{\pi}{4} \; and\; \tan ^{-1}(0)=0$
 
Hence $I=\large\frac{\pi}{4}-0$
 
Hence $ \int \limits_0^1 \large \frac{dx}{1+x^2}=\frac{\pi}{4}$

 

answered Feb 11, 2013 by meena.p
 
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