Guest

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
12 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


Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly.
 We are all IITians and here to help you in your IIT JEE preparation.

 All the best.
 
Regards,
Askiitians Experts
Badiuddin

           


Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free