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The value of $\int\limits_{0}^{1}x(1-x)^{4}dx$ is

\[\begin{array}{1 1}(1)\frac{1}{12}&(2)\frac{1}{30}\\(3)\frac{1}{24}&(4)\frac{1}{20}\end{array}\]

Can you answer this question?
 
 

1 Answer

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$\int \limits_0^1 x(1-x)^4dx= \int \limits_0^1 (1-x)[1-(1-x)]^4 dx$
$\qquad=\int _0^1 (1-x)[1-1+x]^4$$dx$
$\qquad= \int \limits (1-x)x^4 dx$
$\qquad= \int \limits_0^1 (x^4-x^5)dx$
$\qquad= \bigg[\large\frac{x^5}{5} -\frac{x^6}{6}\bigg]_0^1$
$\qquad = \bigg[ \large\frac{1}{5}-0 \bigg] -\bigg[ \large\frac{1}{6} -0 \bigg]$
$\qquad= \large\frac{6-5}{30}=\frac{1}{30}$
Hence 2 is the correct answer.
answered May 21, 2014 by meena.p
 
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