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Home  >>  CBSE XII  >>  Math  >>  Integrals
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By using the properties of definite integrals,evaluate the integral\[\int\limits_0^1x\;(1-x)^n\;dx\]

$\begin{array}{1 1} \frac{1}{(n+1)(n+2)} \\ \frac{1}{(n-1)(n-2)} \\ \frac{n+1}{n+2} \\ \frac{n}{(n+1)(n+2)} \end{array} $

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1 Answer

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Toolbox:
  • (i)$ \int \limits_a^b f(x)dx=F(b)-F(a)$
  • (ii)$\int \limits_0^a f(x)dx=\int \limits_0^a f(a-x)dx$
$\int\limits_0^1x\;(1-x)^n\;dx$
By property (ii)$\int \limits_0^a f(x)dx=\int \limits_0^a f(a-x)dx$
$\int\limits_0^1\;x(1-x)^n\;dx$ can be written as
$\int\limits_0^1\;(1-x)[(1-(1-x)]^n\;dx$
$\Rightarrow\:\int\limits_0^1\;(1-x)(x^n)\;dx$
On seperating the terms
$\Rightarrow\:\int\limits_0^1(x^n-x^{n+1})\;dx$
On integrating we get,
$\bigg[\frac{x^{n+1}}{n+1}\bigg]_0^1-\bigg[\frac{x^{n+2}}{n+2}\bigg]_0^1$
Appllying the limits we get,
$\bigg[\frac{1^{n+1}}{n+1}\bigg]-\bigg[\frac{1^{n+2}}{n+2}\bigg]$
$=\frac{1}{n+1}-\frac{1}{n+2}$
On simplifying $\frac{n+2-n-1}{(n+1)(n+2)}=\frac{1}{(n+1)(n+2)}$
=$\frac{1}{(n+1)(n+2)}$
answered Feb 13, 2013 by meena.p
edited Feb 6, 2014 by rvidyagovindarajan_1
 

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