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Math Help - I've forgotten how to evaluate these

  1. #1
    Super Member angel.white's Avatar
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    I've forgotten how to evaluate these



    \int_{-1}^2 \frac 3x~dx

    I want to do:
    =3ln|a| ]_{-1}^2

    =3ln|2|-3ln|-1| ~~~~~=~~~~~ 3ln(2)-3ln(1) ~~~~~=~~~~~ 3ln(2) ~~~~~=~~~~~ln(8)
    But Function calculator tells me its about 7.68 while ln(8) is about 2.08

    Anyone offer remedial help?
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by angel.white View Post


    \int_{-1}^2 \frac 3x~dx

    I want to do:
    =3ln|a| ]_{-1}^2

    =3ln|2|-3ln|-1| ~~~~~=~~~~~ 3ln(2)-3ln(1) ~~~~~=~~~~~ 3ln(2) ~~~~~=~~~~~ln(8)
    But Function calculator tells me its about 7.68 while ln(8) is about 2.08

    Anyone offer remedial help?
    there is a discontinuity at 0. you must use improper integrals to do this, since the interval has a discontinuity in it

    do you remember how to do that, or should i be more specific?


    listen to you! "remedial", pfft
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  3. #3
    Super Member angel.white's Avatar
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    Quote Originally Posted by Jhevon View Post
    there is a discontinuity at 0. you must use improper integrals to do this, since the interval has a discontinuity in it

    do you remember how to do that, or should i be more specific?


    listen to you! "remedial", pfft
    You're going to have to help me out more, I'm not even sure if I've learned improper integrals :/
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by angel.white View Post
    You're going to have to help me out more, I'm not even sure if I've learned improper integrals :/
    we have improper integrals when the limits of our integrals are not nice. and by not nice, i mean it makes no sense to plug them into our original integrand, or sometimes, even in our integral for that matter. something like \int_0^\infty \sin x~dx is an improper integral. why? because it makes no sense to plug in \infty into something. \infty is not a real number. (that's why whenever we talk about \infty we talk about the limit as we approach \infty and not \infty itself). an integral like \int_0^1 \frac 1x~dx is also improper. why? because \frac 1x is not defined at 0, how can we be evaluating an integral there. that's crazy talk!


    the method of improper integrals allow us to deal with such problems. how? well, the way we are familiar with. how do we usually try to evaluate something at a point we can't plug in? we use limits!

    so, here's how it works. we replace the bad limit point with a variable. i think "N" is the usual one, but it can be whatever. and then we use limits to evaluate the integral at that point.

    example, going back to \int_0^\infty \sin x~dx. we would write this as \lim_{N \to \infty} \int_0^N \sin x~dx

    now we evaluate the integral using N in place of \infty, then we plug in the limits and take the limit as N \to \infty.

    now, not all improper integrals exist. we can only get an answer if we get definite answer for the limit of the integral. otherwise, we say the improper integral "diverges" or "does not exist" whichever's more appropriate.

    so for you problem:

    \int_{-1}^2 \frac 3x~dx = \int_{-1}^0 \frac 3x~dx + \int_0^2 \frac 3x~dx

    = \lim_{N \to 0} \int_{-1}^N \frac 3x~dx + \lim_{A \to 0} \int_A^2 \frac 3x~dx

    now evaluate both integrals and take the limits.
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  5. #5
    Super Member angel.white's Avatar
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    Quote Originally Posted by Jhevon View Post
    we have improper integrals when the limits of our integrals are not nice. and by not nice, i mean it makes no sense to plug them into our original integrand, or sometimes, even in our integral for that matter. something like \int_0^\infty \sin x~dx is an improper integral. why? because it makes no sense to plug in \infty into something. \infty is not a real number. (that's why whenever we talk about \infty we talk about the limit as we approach \infty and not \infty itself). an integral like \int_0^1 \frac 1x~dx is also improper. why? because \frac 1x is not defined at 0, how can we be evaluating an integral there. that's crazy talk!


    the method of improper integrals allow us to deal with such problems. how? well, the way we are familiar with. how do we usually try to evaluate something at a point we can't plug in? we use limits!

    so, here's how it works. we replace the bad limit point with a variable. i think "N" is the usual one, but it can be whatever. and then we use limits to evaluate the integral at that point.

    example, going back to \int_0^\infty \sin x~dx. we would write this as \lim_{N \to \infty} \int_0^N \sin x~dx

    now we evaluate the integral using N in place of \infty, then we plug in the limits and take the limit as N \to \infty.

    now, not all improper integrals exist. we can only get an answer if we get definite answer for the limit of the integral. otherwise, we say the improper integral "diverges" or "does not exist" whichever's more appropriate.

    so for you problem:

    \int_{-1}^2 \frac 3x~dx = \int_{-1}^0 \frac 3x~dx + \int_0^2 \frac 3x~dx

    = \lim_{N \to 0} \int_{-1}^N \frac 3x~dx + \lim_{A \to 0} \int_A^2 \frac 3x~dx

    now evaluate both integrals and take the limits.
    Makes sense.

    \int_{-1}^2\frac 1x ~dx

    =\lim_{N\to 0}\int_{-1}^N\frac 3x ~dx+\lim_{A\to 0}\int_A^2\frac 3x~dx

    =\lim_{N\to 0}[3ln|x|]_{-1}^N +\lim_{A\to 0}[3ln|x|]_{A}^{2}

    =\lim_{N\to\infty}[3ln|N|-3ln|-1|] ~~~~+~~~~\lim_{A\to 0}[3ln|2|-3ln|A|]

    =\lim_{N\to 0}[~~\lim_{A\to 0}(~~3ln|N|~~+~~3ln|2|~~-~~3ln|A|~~)]

    =\lim_{N\to 0}\left[~~\lim_{A\to 0}\left(3ln\left|\frac{2N}{A}\right|~~\right)\righ  t]

    Okay, I'm not sure what to do again, it looks like they cancel out and I'm left with 3ln|2|
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by angel.white View Post
    Makes sense.

    \int_{-1}^2\frac 1x ~dx

    =\lim_{N\to 0}\int_{-1}^N\frac 3x ~dx+\lim_{A\to 0}\int_A^2\frac 3x~dx

    =\lim_{N\to 0}[3ln|x|]_{-1}^N +\lim_{A\to 0}[3ln|x|]_{A}^{2}

    =\lim_{N\to\infty}[3ln|N|-3ln|-1|] ~~~~+~~~~\lim_{A\to 0}[3ln|2|-3ln|A|]

    =\lim_{N\to 0}[~~\lim_{A\to 0}(~~3ln|N|~~+~~3ln|2|~~-~~3ln|A|~~)]

    =\lim_{N\to 0}\left[~~\lim_{A\to 0}\left(3ln\left|\frac{2N}{A}\right|~~\right)\righ  t]

    Okay, I'm not sure what to do again, it looks like they cancel out and I'm left with 3ln|2|
    i forgot to mention. when we split it up into several integrals. we work on them separately. if one diverges, then they all diverge. that is the case here. forget what Function Calculator told you (thanks for the link by the way, it'll come in handy some day), technology usually does not do well with this sort of thing. maple told me it's undefined, so that's on my side, while graph thinks it's a negative answer, which makes no sense.

    the integral is divergent here

    ...or maybe... nah, forget it
    Last edited by Jhevon; February 19th 2008 at 10:37 AM.
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  7. #7
    Super Member angel.white's Avatar
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    Okay, I have a question. I graphed it and it looks like the graph of 1/x (was kind of dumb on my part). Can I not say, then that -1 to 1 will cancel eachother out, and I am left with just the integral from 1 to 2 of ln(x)?

    Similar sort of concept as we used in this problem: http://www.mathhelpforum.com/math-he...-integral.html
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  8. #8
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by angel.white View Post
    Okay, I have a question. I graphed it and it looks like the graph of 1/x (was kind of dumb on my part). Can I not say, then that -1 to 1 will cancel eachother out, and I am left with just the integral from 1 to 2 of ln(x)?

    Similar sort of concept as we used in this problem: http://www.mathhelpforum.com/math-he...-integral.html
    yes, i was thinking about that. that's what the "...or maybe...nah, forget it" part of my last post was about. it seems to make sense to me, but according to the method, it doesn't work that way. but i don't know...
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  9. #9
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    Quote Originally Posted by angel.white View Post
    Okay, I have a question. I graphed it and it looks like the graph of 1/x (was kind of dumb on my part). Can I not say, then that -1 to 1 will cancel eachother out, and I am left with just the integral from 1 to 2 of ln(x)?

    Similar sort of concept as we used in this problem: http://www.mathhelpforum.com/math-he...-integral.html
    You cannot say that \int_{-1}^1 \frac{dx}{x} =0 because it is an odd function. Because we need to know that 1/x is continous on [0,1].
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    ... we need to know that 1/x is continuous on [0,1].
    ah, ok. i don't recall that condition. does it show up in the "proof" that the integral of odd functions over balanced intervals is zero? is it in our text?
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    Quote Originally Posted by Jhevon View Post
    ah, ok. i don't recall that condition. does it show up in the "proof" that the integral of odd functions over balanced intervals is zero? is it in our text?
    Suppose f is intregrable on [0,a] for a>0. If we define g(x) = f(x) if x in [0,a] and g(x) = -f(-x) for x in [-a,0] then the extended function g is integrable on [-a,a] and its integral is zero.

    It is an easy excercise (look at Folland on Section 4.1).
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  12. #12
    Super Member angel.white's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    You cannot say that \int_{-1}^1 \frac{dx}{x} =0 because it is an odd function. Because we need to know that 1/x is continous on [0,1].
    So you agree that it is divergent, then?
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  13. #13
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    Quote Originally Posted by angel.white View Post
    So you agree that it is divergent, then?
    The integral is not defined using Darboux integration (Riemann). Because we require that the function be bounded on the interval in this approach.

    However, there is a way to do this integral. Since it does not fall under this definition we define \int_{-1}^1 \frac{dx}{x} = \int_{-1}^{0^-} \frac{dx}{x} + \int_{0^+}^1 \frac{dx}{x}.

    Now,
    \int_{-1}^{0^-} \frac{dx}{x} = \ln |t| - \ln |1| as t\to 0^- this limit just does not exists. So it is divergent.
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