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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Evaluate $\int \limits_0^ 1 x(1-x)^n dx$

$\begin{array}{1 1} (A) I=\bigg(\large\frac{1}{n+1}-\frac{1}{n+2}\bigg) \\ (B) I=\bigg(\large\frac{1}{n-1}-\frac{1}{n-2}\bigg) \\ (C) I=\bigg(\large\frac{1}{n+2}-\frac{1}{n+1}\bigg) \\ (D) I=\bigg(\large\frac{1}{n+1}+\frac{1}{n+2}\bigg) \end{array} $

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1 Answer

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Toolbox:
  • $ \int \limits_a^b f(x)dx =F(b)-F(a)$
  • $ \int x^n dx= \large\frac{x^{n+1}}{n+1}$
Given $I= \int \limits_0^1 x(1-x)^n dx$
Let $1-x=t, => x =1-t$
on differentiating w.r.t x we get,
$-dx=dt=>dx=-dt$
The limit also changes when we substitute t,
when $x=0,t=1$
$x=1,t=0$
Therefore $ I= \int \limits_1^0 (1-t) t^n (-dt)$
$=-\int \limits_1^0 (1-t) t^n.dt$
we can remove the negative sign by interchanging the limits,
Therefore $I= \int \limits_0^1(1-t)t^n.dt$
on expanding we get
$I= \int \limits_0^1 (t ^n-t^{n+1})dt$
on integrating the terms seperately,
$I= \bigg[ \large\frac{t^{n+1}}{n+1}\bigg]_0^1- \bigg[\frac{t ^{n+2}}{n+2}\bigg]_0^1$
on applying limits
$I=\bigg[\large\frac{1}{n+1}-$$0\bigg]-\bigg[\large\frac{1}{n+2}-$$0\bigg]$
$I=\bigg(\large\frac{1}{n+1}-\frac{1}{n+2}\bigg)$
answered May 6, 2013 by meena.p
 
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