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Improper Integral

Mathematics
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|dw:1350067985984:dw|
I have the solution, looking for help to understand the process
lol .. "Improper Integral" is not quite the term you describe it.

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Other answers:

break x^4 in (x^2)^2 then use substitution
|dw:1350068067914:dw|
yes
are you familiar with last three special integrals formula
three special integral formula? not sure which one you are referring to
did you use double substitution
let x^2 = u sub ... this would end up into inverse typerbolic function.
|dw:1350068248916:dw|these two?
well ... you can try that. but I'll give you a shortcut.
ok I will love to see the shortcut
look at the the inverse hyperbolic of sine http://en.wikipedia.org/wiki/Inverse_hyperbolic_function
@experimentX me to please
|dw:1350068353134:dw|
@experimentX nice one
I see it is the first one
|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.
|dw:1350068751083:dw| solution manual has this as a second step
hmm ... it really was improper intgral!!
Is this because of the square root function's domain being continuous on [a, infinity)?
|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 \[2x dx=\sec^2( \theta) d \theta \]
this doesn't converge.
ok I am following you so far
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)
\[2x dx=\sec^2(\theta) d \theta = > x dx =\frac{1}{2} \sec^2(\theta) d \theta \] Now you go to your little expression and make your "suby's" happen.
|dw:1350068915623:dw|
|dw:1350069081160:dw|
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
@precal I'm not sure what you are asking
In the definition of Improper integrals over infinite intervals I have 3 definitions
It is improper integral because of that infinity thing.
probably with definition of improper integral ... when you have limit goes to infinity ... you have improper integral of first kind.
|dw:1350069479305:dw| provided the limit exists
I should mention that if f is continuous on [a,infinity), then for the above first definition
2nd definition If f is continuous on (-infinty, b], then|dw:1350069615971:dw| provided the limit exists.
|dw:1350069732672:dw| this is called improper integral of First Kind.
3rd definition If f is continuous on (-infinity, infinity), then |dw:1350069715530:dw| probided both limits exist, where c is any real number
|dw:1350069792525:dw|
|dw:1350069861584:dw|
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.
|dw:1350070134924:dw|maybe it is because it should have stated the following limits, I think I see a typo from the source
the second kind appears when you have singularity in domain.|dw:1350070237179:dw|
|dw:1350070363244:dw|
now I see the connection, it has to be establish - no wonder you made the statement earlier about it not being improper
|dw:1350070434570:dw|
ok I thinkg I got it, it was the set up that threw me off. I think I understand the solution better now. Thanks
try watching this http://www.youtube.com/watch?v=KhwQKE_tld0
ok thanks, I can use all the help
Thanks, I enjoyed that lecture.

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