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precal Group TitleBest ResponseYou've already chosen the best response.0
dw:1350067985984:dw
 one year ago

precal Group TitleBest ResponseYou've already chosen the best response.0
I have the solution, looking for help to understand the process
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.3
lol .. "Improper Integral" is not quite the term you describe it.
 one year ago

ipm1988 Group TitleBest ResponseYou've already chosen the best response.0
break x^4 in (x^2)^2 then use substitution
 one year ago

precal Group TitleBest ResponseYou've already chosen the best response.0
dw:1350068067914:dw
 one year ago

ipm1988 Group TitleBest ResponseYou've already chosen the best response.0
are you familiar with last three special integrals formula
 one year ago

precal Group TitleBest ResponseYou've already chosen the best response.0
three special integral formula? not sure which one you are referring to
 one year ago

ipm1988 Group TitleBest ResponseYou've already chosen the best response.0
did you use double substitution
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.3
let x^2 = u sub ... this would end up into inverse typerbolic function.
 one year ago

precal Group TitleBest ResponseYou've already chosen the best response.0
dw:1350068248916:dwthese two?
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.3
well ... you can try that. but I'll give you a shortcut.
 one year ago

precal Group TitleBest ResponseYou've already chosen the best response.0
ok I will love to see the shortcut
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.3
look at the the inverse hyperbolic of sine http://en.wikipedia.org/wiki/Inverse_hyperbolic_function
 one year ago

ipm1988 Group TitleBest ResponseYou've already chosen the best response.0
@experimentX me to please
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.3
dw:1350068353134:dw
 one year ago

ipm1988 Group TitleBest ResponseYou've already chosen the best response.0
@experimentX nice one
 one year ago

precal Group TitleBest ResponseYou've already chosen the best response.0
I see it is the first one
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.3
dw:1350068400466:dw probably you would get inverse hyperbolic for this types ... but generally formula is given as dw:1350068503882:dw ... lol all these forms are inverse hyperbolic function. people usually don't use these.
 one year ago

precal Group TitleBest ResponseYou've already chosen the best response.0
dw:1350068751083:dw solution manual has this as a second step
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.3
hmm ... it really was improper intgral!!
 one year ago

precal Group TitleBest ResponseYou've already chosen the best response.0
Is this because of the square root function's domain being continuous on [a, infinity)?
 one year ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
dw:1350068740846:dw Start with a right triangle Label the sides such that the inside of that radical you have above appears somewhere. dw:1350068780160:dw So we will use substitution \[x^2=\tan( \theta) \] dw:1350068798338:dw Now we need to take derivative of both sides of our substitution \[2x dx=\sec^2( \theta) d \theta \]
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.3
this doesn't converge.
 one year ago

precal Group TitleBest ResponseYou've already chosen the best response.0
ok I am following you so far
 one year ago

precal Group TitleBest ResponseYou've already chosen the best response.0
yes in the end, I know this diverges. I know the final answer, just trying to learn the process or understand the process (is a better description)
 one year ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
\[2x dx=\sec^2(\theta) d \theta = > x dx =\frac{1}{2} \sec^2(\theta) d \theta \] Now you go to your little expression and make your "suby's" happen.
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.3
dw:1350068915623:dw
 one year ago

precal Group TitleBest ResponseYou've already chosen the best response.0
dw:1350069081160:dw
 one year ago

precal Group TitleBest ResponseYou've already chosen the best response.0
I guess I am not sure in the set up why [a, infinity) was used vs (infinity, b] or (infinity, infinity) is it because of the domain of the square root function
 one year ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
@precal I'm not sure what you are asking
 one year ago

precal Group TitleBest ResponseYou've already chosen the best response.0
In the definition of Improper integrals over infinite intervals I have 3 definitions
 one year ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
It is improper integral because of that infinity thing.
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.3
probably with definition of improper integral ... when you have limit goes to infinity ... you have improper integral of first kind.
 one year ago

precal Group TitleBest ResponseYou've already chosen the best response.0
dw:1350069479305:dw provided the limit exists
 one year ago

precal Group TitleBest ResponseYou've already chosen the best response.0
I should mention that if f is continuous on [a,infinity), then for the above first definition
 one year ago

precal Group TitleBest ResponseYou've already chosen the best response.0
2nd definition If f is continuous on (infinty, b], thendw:1350069615971:dw provided the limit exists.
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.3
dw:1350069732672:dw this is called improper integral of First Kind.
 one year ago

precal Group TitleBest ResponseYou've already chosen the best response.0
3rd definition If f is continuous on (infinity, infinity), then dw:1350069715530:dw probided both limits exist, where c is any real number
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.3
dw:1350069792525:dw
 one year ago

precal Group TitleBest ResponseYou've already chosen the best response.0
dw:1350069861584:dw
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.3
as long as you have infinity as limits ... just relax .. they all are of same type of improper integral. probably you are confused with improper integral of second kind.
 one year ago

precal Group TitleBest ResponseYou've already chosen the best response.0
dw:1350070134924:dwmaybe it is because it should have stated the following limits, I think I see a typo from the source
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.3
the second kind appears when you have singularity in domain.dw:1350070237179:dw
 one year ago

precal Group TitleBest ResponseYou've already chosen the best response.0
dw:1350070363244:dw
 one year ago

precal Group TitleBest ResponseYou've already chosen the best response.0
now I see the connection, it has to be establish  no wonder you made the statement earlier about it not being improper
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.3
dw:1350070434570:dw
 one year ago

precal Group TitleBest ResponseYou've already chosen the best response.0
ok I thinkg I got it, it was the set up that threw me off. I think I understand the solution better now. Thanks
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.3
try watching this http://www.youtube.com/watch?v=KhwQKE_tld0
 one year ago

precal Group TitleBest ResponseYou've already chosen the best response.0
ok thanks, I can use all the help
 one year ago

precal Group TitleBest ResponseYou've already chosen the best response.0
Thanks, I enjoyed that lecture.
 one year ago
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