Evaluate :
           \int_{0}^{1}x\left ( 1-x \right )^{n}dx

 

 

 

 
 
 
 
 

Answers (1)

I= \int_{0}^{1}x\left ( 1-x \right )^{n}dx                               1-x= t
                                                                        -dx= dt
I= -\int_{1}^{0}\left ( 1-t \right )t^{n}dt
I= \int_{0}^{1}\left ( t^{n}-t^{n+1} \right )dt\: \: = \left [ \frac{t^{n+1}}{n+1}\: \: \frac{-t^{n+2}}{n+2} \right ]^{1}_{0}
I=\left ( \frac{1}{n+1}-\frac{1}{n+2} \right )= \frac{n+2-n-1}{\left ( n+1 \right )\left ( n+2 \right )}= \frac{1}{\left ( n+1 \right )\left ( n+2 \right )}
I=\frac{1}{\left ( n+1 \right )\left ( n+2 \right )}

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