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#math
<adamjon858> hey
<asphyxia> As I read it.
<adamjon858> any calc gurus in here?
<asphyxia> adamjon858: Hi. Perhaps. Try stating your problem, and see if anyone can help you out
<adamjon858> I've got a bit of a trouble with a problem here...I'll type it up..but how should I put in the integral sign?
<wolfbone> asphyxia, transpose x1 and a, transpose x2 and a, transpose x1 and a
<asphyxia> wolfbone: Ah, okay
<adamjon858> $ = integral. $ x^4 lnx dx
<asphyxia> adamjon858: You need to solve that? Try with partial integration
<adamjon858> Should I use integration by parts or substitution?
<asphyxia> parts.
<adamjon858> I tried integration by parts and I get stuck.
<asphyxia> Let f(x) := x^4 and g(x) = lnx
<adamjon858> hrm
<asphyxia> Then you have from the integration by parts, that $ x^4 lnx dx = F(x)g(x) - $ F(x)g'(x)dx
<asphyxia> I can tell you that I get 1/5x^5(lnx - 1/5)
<adamjon858> ah..I used lnx for g(x)
<adamjon858> whoops
<adamjon858> I meant f(x)
<adamjon858> Actually, My textbook uses u and v
<adamjon858> so i had u=lnx, du=dx/x, dv=x^4, v=(x^5)/5
<asphyxia> well, yeah, you can do that to. It gets really simple as (ln(x))' = 1/x
<adamjon858> hrm...I'm really kind of mixed up
<adamjon858> I have $udv = vdu - $uv
<dn4> Does anyone know of a good way to understand why lim x->0 sin(x)/(x) = 1
<adamjon858> Plugging this in I get $lnxx^4dx = x^5/5 * lnx - $ x^5/5 dx/x
<asphyxia> dn4: It is solved by using L'Hopital's theorem
<adamjon858> and now I'm comletely lost
<Dacicus> adamjon858: I think the proof uses the sandwich rule
<Dacicus> or whatever name you learned it a
<adamjon858> so now I'm stuck with $ 1/5 x^5 1/x dx
<Dacicus> as*
<adamjon858> sandwich rule?
<asphyxia> dn4: Which states that if both denominator and numerator approaches 0, then f/g = f'/g'
<adamjon858> I'm really lost...I feel so dumb
<asphyxia> adamjon858: $ 1/5 x^5 1/x dx
<asphyxia> that is easy to simplify
<adamjon858> hrm
<asphyxia> think of x^4/x instead
<adamjon858> asphyxia: prove it
<asphyxia> oh
<asphyxia> x^5/x
<adamjon858> ah because I can take the 1/5 out right
<asphyxia> 1/5 * x^5 * 1/x = 1/5 * x^5/x = 1/5 * x^4.
<cyclicFifths> adamjon858: what is 16/4 ? ie 2^4 / 2^2 ?
<adamjon858> i see
<adamjon858> I feel dumb
<cyclicFifths> don't feel dumb, just work problems :)
<asphyxia> dn4: So that's why, because when x->0, sin(x) -> 0 and (sin(x))' = cos(x) -> 1. Likewise x->0 and therefore x' = x -> 1.
<adamjon858> so the answer to that particular part of the integral would be x^5/25
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