Prove that \int_{0}^{a}f\left ( x \right )dx= \int_{0}^{a}f\left ( a-x \right )dx,   and  hence evaluate \int_{0}^{1}x^{2}\left ( 1-x \right )^{n}dx.

 

 

 

 
 
 
 
 

Answers (1)

\int_{0}^{a}f\left ( x \right )dx= \int_{0}^{a}f\left ( a-x \right )dx,\left [ given \right ]
Let x= a-t\Rightarrow dx= -dt\: also\: when\: x= 0\Rightarrow t= a
& when x= a\Rightarrow t= 0
so \int_{0}^{a}f\left ( x \right )dx= \int_{a}^{0}f\left ( a-t \right )\left ( -dt \right )= \left [ -\int_{a}^{0}f\left ( a-t \right )dt \right ]
                                                                            = \int_{0}^{a}f\left ( a-t \right )dt
replacing t by x
Hence = \int_{0}^{a}f\left ( x \right )dx= \int_{0}^{a}f\left ( a-x \right )dx  
Let\, I= \int_{0}^{1}x^{2}\left ( 1-x \right )^{n}dx
           = \int_{0}^{1}\left ( 1-x \right )^{2}\left \{ 1-\left ( 1-x \right ) \right \}^{n}dx
          = \int_{0}^{1}\left ( 1-x \right )^{2}x^{n}dx
         = \int_{0}^{1}\left ( x^{n}-2x^{1+n}+x^{2+n} \right )dx
        = \left [ \frac{x^{n+1}}{n+1}-\frac{2x\, x^{n+2}}{n+2}+\frac{x^{n+3}}{n+3} \right ]^{1}_{0}
    I= \left [ \frac{1}{n+1}-\frac{2}{n+2}+\frac{1}{n+3} \right ]-\left [ 0-2\times 0+0 \right ]
       I= \frac{1}{n+3}-\frac{n}{n^{2}+3n+2}
         = \frac{n^{2}+3n+2-n^{2}-3n}{\left ( n+1 \right )\left ( n+2 \right )\left ( n+3 \right )}
\therefore I= \frac{2}{\left ( n+1 \right )\left ( n+2 \right )\left ( n+3 \right )}

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