If I(m,n)=\int_{0}^{1}{t^m(1+t)^ndt}, then the expression for I(m,n) in terms of I(m+1,n-1) is:
a)2nm+1-nm+1I(m+1,n-1)
b)nm+1I(m+1,n-1)
c)2nm+1+nm+1I(m+1,n-1)
d)mn+1I(m+1,n-1)
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3 Answers
1In these type of problems , the aim is to part - integrate the integral in such a way that the second function is the one whose " value " , i . e , in this case , the power of " t m " is seen to have increased . In the solution below , the limits are from " 0 " to " 1 " .
I ( m , n ) = ∫ t m ( 1 + t ) n dt
= ( 1 + t ) n ∫ t m dt - ∫ d ( 1 + t ) ndt [ ∫ t m dt ] dt
= [ t m + 1 ( 1 + t ) nm + 1 ] - nm + 1 ∫ t m + 1 ( 1 + t ) n - 1 dt
= 2 nm + 1 - nm + 1 I ( m + 1 , n - 1 )........................After putting the limits .
Hence , the answer is " ( a ) " .
39@Ricky - you used 0, 1 as the limits because the integral is strikingly similar to a beta function, isn't it? :P
49@Pritish: No..the question provides the limit! ;)
@Ricky: Thank you bhai! :D