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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Evaluate the following: $\int\limits_0^1\frac{xdx}{\sqrt{1+x^2}}$

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Toolbox:
  • $\int \limits_a^b x^n=\bigg[ \large \frac{x^{n+1}}{n+1}\bigg]_a^b$
Let $I=\large\int\limits_0^1\frac{xdx}{\sqrt{1+x^2}}$
Put $1+x^2=t$
On differentiating with respect to t we get,
$2xdx=dt =>xdx=dt/2$
When we substitute for t, the limits also change
When $x=0, \qquad 1+0=t=>t=1$
When $x=1, \qquad 1+1=t=>t=2$
On substituting this in I we get,
Therefore $I=\large\frac{1}{2} \int \limits^2_1 \frac{dt}{\sqrt t}$
$=\large\frac{1}{2} \int \limits _1^2 t^{\frac{-1}{2}}$$dt$
On integrating we get
$I= \large \frac{1}{2} \bigg[\frac{t ^{1/2}}{1/2} \bigg]_1^2$
$=\bigg[ \sqrt t \bigg]^2_1$
On applying limits we get
$I=\sqrt 2-1$

 

answered Apr 18, 2013 by meena.p
edited Apr 18, 2013 by meena.p
 
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