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for m>0,n>0.show integration ---x(pow)m (1-x)(Pow)n from 0 to 1= m!n!/(m+n+1)!

for m>0,n>0.show integration ---x(pow)m (1-x)(Pow)n from 0 to 1=


m!n!/(m+n+1)!

Grade:12

1 Answers

Badiuddin askIITians.ismu Expert
148 Points
14 years ago

Dear yashika

for m>0,n>0

Im,n=∫xm(1-x)n dx  limit 0 to 1

        intigrate by using 1st and 2nd function

       =-1/(n+1)  xm(1-x)n+1  limit 0 to 1  +m/(n+1) ∫xm-1(1-x)n+1 dx  limit 0 to 1

       =0 +m/(n+1) ∫xm-1(1-x)(1-x)n dx  limit 0 to 1

       = m/(n+1) [∫xm-1(1-x)n dx - ∫xm(1-x)ndx ]    limit 0 to 1

       =m/(n+1) [Im-1,n  - Im,n   ]

simplify     Im,n   =   m/(m+n+1) Im-1,n 

                       =m*(m-1)/(m+n+1)(m+n)  Im-2,n

  repeat this process you will get desire result


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