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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Evaluate:$\int \limits_0^ 1 \Large \frac{dx}{\sqrt{x+1}+\sqrt x}$

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Toolbox:
  • $(i)\; \int \limits_a^b f(x)dx =F(b)-F(a)$
  • $(ii)\; \int x^n dx =\large\frac{x^{n+1}}{n+1}$
Given $I=\int \limits_0^ 1 \large\frac{dx}{\sqrt{x+1}+\sqrt x}$
Let us multiply and divide by the conjugate $\sqrt {x+1}-\sqrt x$
Therefore $ I= \int \limits_0^1 \large\frac{\sqrt {x+1} -\sqrt x}{(\sqrt {x+1}+\sqrt x)(\sqrt {x+1}-\sqrt x)}dx$
$=\int \limits_0^1 \large\frac{\sqrt {x+1}-\sqrt x}{(\sqrt x+1)^2 -(\sqrt x)^2 } dx$
On simplifying we get
$\int \limits_0^1 \large\frac{\sqrt {x+1}-\sqrt x}{x+1-x}dx$
$=\int \limits_0^1 \sqrt {x+1}-\sqrt x$
On seperating the terms
$\int \limits_0^1 \sqrt {x+1} dx-\int \limits_0^1 \sqrt x dx$
On integrating we get,
$\bigg[ \large\frac{(x+1)^{3/2}}{3/2}\bigg]_0^1-\bigg[\frac {x^{3/2}}{3/2} \bigg]_0^1$
$=\large\frac{2}{3}$$\bigg[(x+1)^{3/2}\bigg]_0^1 -\large\frac{2}{3} $$\bigg[x^{3/2}\bigg]_0^1$
On applying the limits
$\large\frac{2}{3}$$\bigg[(1+1)^{3/2}-(0+1)^{3/2} \bigg]-\large\frac{2}{3} $$\bigg[1^{3/2}-0\bigg]$
$= \large \frac{2}{3} $$\bigg[2^{3/2}-1\bigg]-\large\frac{2}{3}(1)$
$=\large\frac{2}{3}$$\bigg[2 \sqrt 2-2\bigg]$
$=\large\frac{4}{3} $$[ \sqrt 2 -1]$
answered Apr 25, 2013 by meena.p
 
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