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precal
 3 years ago
Improper Integral
precal
 3 years ago
Improper Integral

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precal
 3 years ago
Best ResponseYou've already chosen the best response.0I have the solution, looking for help to understand the process

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.3lol .. "Improper Integral" is not quite the term you describe it.

ipm1988
 3 years ago
Best ResponseYou've already chosen the best response.0break x^4 in (x^2)^2 then use substitution

ipm1988
 3 years ago
Best ResponseYou've already chosen the best response.0are you familiar with last three special integrals formula

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

ipm1988
 3 years ago
Best ResponseYou've already chosen the best response.0did you use double substitution

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

precal
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1350068248916:dwthese two?

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.3well ... you can try that. but I'll give you a shortcut.

precal
 3 years ago
Best ResponseYou've already chosen the best response.0ok I will love to see the shortcut

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

ipm1988
 3 years ago
Best ResponseYou've already chosen the best response.0@experimentX me to please

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.3dw:1350068353134:dw

precal
 3 years ago
Best ResponseYou've already chosen the best response.0I see it is the first one

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.3dw: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.

precal
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1350068751083:dw solution manual has this as a second step

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.3hmm ... it really was improper intgral!!

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

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.1dw: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 \[2x dx=\sec^2( \theta) d \theta \]

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.3this doesn't converge.

precal
 3 years ago
Best ResponseYou've already chosen the best response.0ok I am following you so far

precal
 3 years ago
Best ResponseYou've already chosen the best response.0yes 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)

myininaya
 3 years ago
Best 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.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.3dw:1350068915623:dw

precal
 3 years ago
Best ResponseYou've already chosen the best response.0I 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

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.1@precal I'm not sure what you are asking

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

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.1It is improper integral because of that infinity thing.

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

precal
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1350069479305:dw provided the limit exists

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

precal
 3 years ago
Best ResponseYou've already chosen the best response.02nd definition If f is continuous on (infinty, b], thendw:1350069615971:dw provided the limit exists.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.3dw:1350069732672:dw this is called improper integral of First Kind.

precal
 3 years ago
Best ResponseYou've already chosen the best response.03rd definition If f is continuous on (infinity, infinity), then dw:1350069715530:dw probided both limits exist, where c is any real number

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.3dw:1350069792525:dw

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.3as 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.

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

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

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

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.3dw:1350070434570:dw

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

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

precal
 3 years ago
Best ResponseYou've already chosen the best response.0ok thanks, I can use all the help

precal
 3 years ago
Best ResponseYou've already chosen the best response.0Thanks, I enjoyed that lecture.
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