Results 1 to 2 of 2

Math Help - How to determine the convergence

  1. #1
    Newbie
    Joined
    May 2010
    Posts
    6

    How to determine the convergence

    Hi, I don't know to determine the convergence of these integrals:

    The first one is {\displaystyle \int_{0}^{1}\frac{3\sin^{2}(x)+5\cos^{5}(x)}{\sqrt  {x}(x+1)}}dx

    and the second is {\displaystyle \int_{1}^{\infty}\frac{2+x^{3}}{1+x^{6}}dx}

    \displaystyle

    For the second I tried the following <br /> <br />
{\displaystyle I=\int_{1}^{\infty}\frac{2}{x^{6}(1+\frac{1}{x^{6}  })}dx}+\int_{1}^{\infty}\frac{1}{x^{3}(1+\frac{1}{  x^{6}})}dx=\int_{1}^{\infty}f_{1}(x)dx+\int_{1}^{\  infty}f_{2}(x)dx<br />
    When  \displaystyle x\rightarrow\infty \begin{cases}<br />
x^{6}\rightarrow & \infty\\<br />
(1+\frac{1}{x^{6}}) & \rightarrow1\\<br />
x^{3} & \rightarrow\infty\end{cases}

    Then
    f_{1}(x)\sim\frac{2}{x^{6}} , Thus  I_{1}converges

    {\displaystyle f_{2}(x)}\sim\frac{1}{x_{3}}\Rightarrow I_{2} converges, then  I converges.
    my reasoning is correct?

    Bye and thanks for everything.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Aug 2007
    From
    USA
    Posts
    3,110
    Thanks
    2
    No, that piece-meal convergence idea is not as useful as you want it to be. That's why there is no such theorem. You just need to make sense of it. You're just demonstrating convergence.

    \frac{2+x^{3}}{1+x^{6}}\;<\;\frac{2+x^{3}}{x^{6}}

    Now, it's pretty obvious. I just made the denominator a little smaller.

    Let's just play with the other one and see if it leads anywhere...

    With only a little thought:

    \frac{-\sqrt{34}}{\sqrt{x}(x+1)}\;<\;\frac{3\sin^{2}(x) + 5\cos^{5}(x)}{\sqrt{x}(x+1)}\;<\;\frac{\sqrt{34}}{  \sqrt{x}(x+1)}

    We could narrow it down more than that, but that may suffice.

    \frac{-\sqrt{34}}{\sqrt{x+1}(x+1)}\;<\;\frac{-\sqrt{34}}{\sqrt{x}(x+1)}\;<\;\frac{3\sin^{2}(x) + 5\cos^{5}(x)}{\sqrt{x}(x+1)}\;<\;\frac{\sqrt{34}}{  \sqrt{x}(x+1)}\;<\;\frac{\sqrt{34}}{\sqrt{x}(x)}

    What say you? Did that get us anywhere? Actually, no, but let's see what you can present.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. determine the convergence
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 13th 2011, 01:52 PM
  2. Determine convergence or divergence
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 17th 2010, 06:46 PM
  3. Determine values for convergence
    Posted in the Calculus Forum
    Replies: 2
    Last Post: November 24th 2009, 08:14 AM
  4. determine interval of convergence
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 12th 2009, 11:37 PM
  5. Determine the convergence.
    Posted in the Calculus Forum
    Replies: 5
    Last Post: March 14th 2008, 01:21 PM

Search Tags


/mathhelpforum @mathhelpforum