Mental block on antiderrivating the integral from one to infinity of 1/(square root of x+1).
Thanks!
this is an improper integral, so we have to use limits when evaluating it. for the anti derivative, we use substitution.
int{0-->infinity} [1/sqrt(x + 1)]dx
= int{0-->infinity} [(x + 1)^(-1/2)]dx
let u = x + 1
=> du = dx
so our integral becomes:
int{u^(-1/2)}du
= 2u^(1/2) + C
= 2(x + 1)^(1/2) + C
now to evaluate between 0 and infinity, we proceed like this
int{0-->infinity} [(x + 1)^(-1/2)]dx = lim{N-->infinity} int{0-->N} [(x + 1)^(-1/2)]dx
= lim{N-->infinity} 2(x + 1)^(1/2) evaluate between N and 0
= lim{N-->infinity} 2(N + 1)^(1/2) - 2(0 + 1)^(1/2)
= lim{N-->infinity} 2(N + 1)^(1/2) - 2
= infinity
Oh yeah, that's right. the limits of integration were from 1 to infinity. i did for 0 to infinity. the answer is pretty much the same though, except for the third to last line, you plug in 1 instead of 0. the answer remains the same, it diverges to infinity.
= lim{N-->infinity} 2(N + 1)^(1/2) - 2(1 + 1)^(1/2)
= lim{N-->infinity} 2(N + 1)^(1/2) - 2sqrt(2)
= infinity
dang! i've got to learn to read again! i've been misreading posts all week. i guess it's because i'm hardly getting any sleep these days, i'm constantly tired--but maybe that's a sorry excuse