# Answer on Calculus Question for Audrey

Question #20394

Use the Beta function to evaluate (1 + x)^m (1 - x)^n dx (-1,1)

Expert's answer

integral (1 + x)^m (1 - x)^n dx on (-1,1) let t=(1+x)/2 then 1+x=2t, x=2t-11-x=1-2t+1=2(1-t)

dt=dx/2 so dx=2dt and after substitution we have

integral (1 + x)^m (1 - x)^n dx on (-1,1)=integral (2t)^m( 2(1-t) )^n 2dt on (0,1)= =integral 2^m t^m

2^n (1-t)^n 2*dt on (0,1)==2^(m+n+1) * integral t^m *(1-t)^n dt on (0,1)=

=2^(m+n+1)*BETTA-function(m+1,n+1)

dt=dx/2 so dx=2dt and after substitution we have

integral (1 + x)^m (1 - x)^n dx on (-1,1)=integral (2t)^m( 2(1-t) )^n 2dt on (0,1)= =integral 2^m t^m

2^n (1-t)^n 2*dt on (0,1)==2^(m+n+1) * integral t^m *(1-t)^n dt on (0,1)=

=2^(m+n+1)*BETTA-function(m+1,n+1)

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