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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class12  >>  Integral Calculus
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Integrate $\int \limits_0^1 x^m (1-x)^n.dx$

\[\begin {array} {1 1} (a)\;I_{m,n}=\frac{n}{m+1} I_{m+n.n-1} \\ (b)\;I_{m,n}=-\frac{n}{m+1} I_{m+n.n-1} \\ (c)\;I_{m,n}=\frac{2^n}{m+1}-\frac{n}{m+1} I_{m+n.n-1} \\ (d)\;I_{m,n}=\frac{2^n}{m+1}+\frac{n}{(m+1)} I_{m+n.n-1} \end {array}\]

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1 Answer

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Integration By $\int uv.dx:$
$I_{m,n}= \bigg[ (1-x)^n. \large\frac{x^{m+1}}{m+1}\bigg]_0^1-\int \limits_0^1 \large\frac{x^{m+1}}{m+1}$$ xn(1-x)^{n-1}.dx$
By putting limits
$\qquad= \bigg[ (1-1)^n. \large\frac{1^{m+1}}{m+1}- \large\frac{(1-0)^n.(0)^{m+1}}{m+1}\bigg]-\int \limits_0^1 \large\frac{n}{m+1}.$$x^{m+1}.(1-x)^{n-1}dx$
$\qquad= -\large\frac{n}{m+1} $$\int \limits_0^1 x^{m+1} (1-x) ^{n-1} dx$
$ I_{m,n}=\large\frac{-n}{m+1} I_{m+n.n-1}$
Hence b is the correct answer.
answered Dec 12, 2013 by meena.p
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