Results 1 to 3 of 3

Math Help - concepts of improper integrals

  1. #1
    Member
    Joined
    Jan 2009
    Posts
    100

    concepts of improper integrals

    \int \frac{10}{x^2-2x}dx [0,3]

    To determine the convergence or divergence of the integral, how many improper integrals must be analyzed? What must be true of each of these integrals if the given integral converges?

    I think it must be analyzed twice. Even if that is right, I don't know what must be true of each of them if they converge.

    Please help.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,621
    Thanks
    426
    Quote Originally Posted by saiyanmx89 View Post
    \int \frac{10}{x^2-2x}dx [0,3]

    To determine the convergence or divergence of the integral, how many improper integrals must be analyzed? What must be true of each of these integrals if the given integral converges?

    I think it must be analyzed twice. Even if that is right, I don't know what must be true of each of them if they converge.

    Please help.

    note that \int_0^3 \frac{10}{x^2-2x}dx has to be broken up into three improper integral "pieces"

    \lim_{a \to 0} \int_a^1 \frac{10}{x(x-2)} \, dx

    \lim_{b \to 2} \int_1^b \frac{10}{x(x-2)} \, dx

    \lim_{b \to 2} \int_b^3 \frac{10}{x(x-2)} \, dx

    If the overall integral converges, then each "piece" of the integral must converge. if any single "piece" diverges, then the entire integral diverges.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor Reckoner's Avatar
    Joined
    May 2008
    From
    Baltimore, MD (USA)
    Posts
    1,024
    Thanks
    75
    Awards
    1

    Smile

    Quote Originally Posted by saiyanmx89 View Post
    \int \frac{10}{x^2-2x}dx [0,3]

    To determine the convergence or divergence of the integral, how many improper integrals must be analyzed? What must be true of each of these integrals if the given integral converges?

    I think it must be analyzed twice. Even if that is right, I don't know what must be true of each of them if they converge.

    Please help.
    The integrand has infinite discontinuities at x=0 and x=2. If the integral does converge, then we may write it as

    \int_0^3\frac{10}{x^2-2x}\,dx

    =\int_0^1\frac{10}{x^2-2x}\,dx+\int_1^2\frac{10}{x^2-2x}\,dx+\int_2^3\frac{10}{x^2-2x}\,dx (I chose to split the first interval at 1, but you could use any value between 0 and 2)

    =\lim_{a\to0^+}\left[\int_a^1\frac{10}{x^2-2x}\,dx\right]+\lim_{b\to2^-}\left[\int_1^b\frac{10}{x^2-2x}\,dx\right]+\lim_{a\to2^+}\left[\int_a^3\frac{10}{x^2-2x}\,dx\right]

    If any of these three limits does not exist, then the original integral does not converge. For the antiderivative, use partial fraction decomposition.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. improper integrals
    Posted in the Calculus Forum
    Replies: 3
    Last Post: June 4th 2011, 11:34 PM
  2. Replies: 3
    Last Post: August 8th 2010, 08:29 PM
  3. Improper Integrals
    Posted in the Calculus Forum
    Replies: 4
    Last Post: April 5th 2010, 01:45 PM
  4. Improper Integrals
    Posted in the Calculus Forum
    Replies: 16
    Last Post: April 19th 2008, 06:43 PM
  5. Improper Integrals
    Posted in the Calculus Forum
    Replies: 2
    Last Post: April 18th 2008, 07:17 PM

Search Tags


/mathhelpforum @mathhelpforum