Subsstitution Question (1 Viewer)

[Kutay]

Hey i got a question not sure with this one i get the answer of

2(1^n+1) - 1^n+2
------------------------
2

the question is a show one and i cant get to the answer of this

Use the substitution u = 1 - x to prove that for n > 0.

1 INTEGRAL 0 x(1 - x)^n = 1
---------------------------
(n+1) (n+2)

i cant get it to show that...... can someone go through the working solutions to show why i cant get to that answer ???

 

[FinalFantasy]

do u mean
I=int. x(1-x)^n dx from x=0 to x=1?

let u=1-x
du=-dx
I=-int. (1-u)u^n du from u=1 to u=0
=int. (u^n-u^(n+1) ) du from u=0 to u=1
=(1\(n+1)) u^(n+1)- (1\(n+2) )u^(n+2)
=[(n+2)u^(n+1)-(n+1)u^(n+2)]\(n+1)(n+2)
from u=0 to u=1

putting dem in
I=[(n+2)-(n+1)]\(n+1)(n+2) -0
=1\(n+1)(n+2)

hope it's not too messy lol

 

[Kutay]

i can see how u done but whats the I thingy its I of between 1 and 0 .hmm anyhow i iwll look over your working see what i have done wrong thankyou.

 

[Sepulchres]

Meh, I did it too. Here.

 

[Kutay]

oh heh i see where i went wrong it all good i do a dif way to you

 

[FinalFantasy]

i can see how u done but whats the I thingy its I of between 1 and 0 .hmm anyhow i iwll look over your working see what i have done wrong thankyou.

what u mean "I thingy"?
i let I=the integral for convenience if i need to write it again later
and int. something from a to b
=- int. something from b to a
u can switch it like dat
if that's wat ur looking for

 

[Kutay]

putting dem in
I=[(n+2)-(n+1)]\(n+1)(n+2) -0
=1\(n+1)(n+2)

i dont understnad how from that line u get to that final line if it is that + that dont u make the bttom the same so it couldn't = that... thats where i am stuck :|

 

[FinalFantasy]

^
numerator: n+2-n-1=1
denominator: (n+1)(n+2)
so it's 1\(n+1)(n+2)
if it's numerator u dun need to "make the bttom the same"
u do it for the denominator
does dat ans. ur q?

 

[Kutay]

yes thankyou very much

 

[acmilan]

Meh, I did it too. Here.

slight error with the signs in the last few lines