Results 1 to 11 of 11

Math Help - integral

  1. #1
    Member
    Joined
    Nov 2006
    Posts
    139

    integral

    the improper integral from 1 to + infinity

    3e^(-x) - x^(4)
    ______________

    2x^(3) + x^(5)

    A)diverges to + inf
    B)exists and is finite
    C)doesn't exist
    D)none of the preceding

    The right answer is D but I can't see why.
    Thank you for your help.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,671
    Thanks
    299
    Awards
    1
    Quote Originally Posted by 0123 View Post
    the improper integral from 1 to + infinity

    3e^(-x) - x^(4)
    ______________

    2x^(3) + x^(5)

    A)diverges to + inf
    B)exists and is finite
    C)doesn't exist
    D)none of the preceding

    The right answer is D but I can't see why.
    Thank you for your help.
    I'm not quite sure what the definition of "doesn't exist" is supposed to be because any integral that is (+/-) infinity doesn't exist, so A and C should be the same!

    Regardless, look at the form of the integrand for large x (which translates in this case for any x greater than 5 - 10 or so.)
    \frac{3e^{-x} - x^4}{2x^3 + x^5} \to -\frac{1}{x}

    So compute \int_{10}^{\infty} dx \left ( - \frac{1}{x} \right ) \to -\infty.

    This by itself is not an answer, but if you sketch the integrand over [1, 10] you will see that the area under the curve is clearly finite. Thus
    \int_1^{\infty}dx \frac{3e^{-x} - x^4}{2x^3 + x^5} \to \text{finite} - \infty \to -\infty

    Since this answer isn't listed, it must be answer D.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by topsquark View Post
    I'm not quite sure what the definition of "doesn't exist" is supposed to be because any integral that is (+/-) infinity doesn't exist, so A and C should be the same!

    Regardless, look at the form of the integrand for large x (which translates in this case for any x greater than 5 - 10 or so.)
    \frac{3e^{-x} - x^4}{2x^3 + x^5} \to -\frac{1}{x}

    So compute \int_{10}^{\infty} dx \left ( - \frac{1}{x} \right ) \to -\infty.

    This by itself is not an answer, but if you sketch the integrand over [1, 10] you will see that the area under the curve is clearly finite. Thus
    \int_1^{\infty}dx \frac{3e^{-x} - x^4}{2x^3 + x^5} \to \text{finite} - \infty \to -\infty

    Since this answer isn't listed, it must be answer D.

    -Dan
    For large x

    \frac{3e^{-x} - x^4}{2x^3 + x^5} = -O(x^{-1})

    which means that for large x there exist positive constants k and K such that:

    -k\, x^{-1}<\frac{3e^{-x} - x^4}{2x^3 + x^5} < -K\,x^{-1}

    Which is sufficient to show that the integral diverges to -\infty, that is the integral does not exist.

    RonL
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Nov 2006
    Posts
    139
    Firstly, thank you both. Then, I am getting a little bit confused..ok, now, I sketched the graph and clearly it shows the area is finite, but without that there are still some points that I don't get (sorry for annoying you.. ) I arrived 'till (-1/x), since qualitatively we can say the integrand will be like that, but then I am lost. As far as I remember, 1/x diverges and if something diverges then the area is infinite...which is surely not correct, otherwise the answer would have been another...help!
    thanks thanks thanks so much
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by 0123 View Post
    Firstly, thank you both. Then, I am getting a little bit confused..ok, now, I sketched the graph and clearly it shows the area is finite
    How does the sketch show the area to be finite?

    Just because the absolute value of the integrand is decreasing as x increases does not make the area finite.

    RonL
    Attached Thumbnails Attached Thumbnails integral-gash.jpg  
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,671
    Thanks
    299
    Awards
    1
    Quote Originally Posted by CaptainBlack View Post
    How does the sketch show the area to be finite?

    Just because the absolute value of the integrand is decreasing as x increases does not make the area finite.

    RonL
    The integral is finite over the interval [1, 10]. That is what I was referring to.

    -Dan

    Edit: The way I replied to this makes it look like I was replying to CB. My response was meant for 0123. Sorry about that!
    Last edited by topsquark; January 10th 2007 at 05:31 AM. Reason: Oops!
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Nov 2006
    Posts
    139
    sorry.. I got it, thanks!!!
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,671
    Thanks
    299
    Awards
    1
    Quote Originally Posted by 0123 View Post
    well the book says "none of the preceding", not "doesn't exist"... and well, what did I(=the program) sketched since I came out with a different graph
    so, now...I don't understand..
    thanks for your help and for your patience
    The point is that both your integrand and -1/x share similar characteristics at large x. This is enough to say that both integrals are infinite. CaptainBlack's graph showed what the integrand looks like for x > 0. My own graph confirms his. I'm not sure what your graph looks like. (Unless you sketched it for all x, in which case it does look rather different. But remember all we need is the interval [1, infinity). )

    -Dan
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by topsquark View Post
    The integral is finite over the interval [1, 10]. That is what I was referring to.

    -Dan

    Edit: The way I replied to this makes it look like I was replying to CB. My response was meant for 0123. Sorry about that!
    Sorry about the confusion, I hadn't realised that 0123 was referring to
    what you wrote (which of course was right)

    RonL
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Member
    Joined
    Nov 2006
    Posts
    139
    sorry also on my part, I created much of this confusion. and thanks again
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,671
    Thanks
    299
    Awards
    1
    Quote Originally Posted by CaptainBlack View Post
    Sorry about the confusion, I hadn't realised that 0123 was referring to
    what you wrote (which of course was right)

    RonL
    (snort) Lately what I've been writing distinctly does NOT fall under the category "which of course was right." I'd say I need a vacation, but I just got off one!

    -Dan
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: August 31st 2010, 07:38 AM
  2. Replies: 1
    Last Post: June 2nd 2010, 02:25 AM
  3. Replies: 0
    Last Post: May 9th 2010, 01:52 PM
  4. Replies: 0
    Last Post: September 10th 2008, 07:53 PM
  5. Replies: 6
    Last Post: May 18th 2008, 06:37 AM

Search Tags


/mathhelpforum @mathhelpforum