Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

imagreencat

  • 3 years ago

We were to asked to get the integral from 0 to infinity of (dx)/(sqrt(x))(x+1). I don't know if I should use the infinite intervals AND the discontinuous integrand or should I just go with one?

  • This Question is Closed
  1. lgbasallote
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    split them up first

  2. lgbasallote
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    1 to infity and 0 to 1 should work

  3. imagreencat
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    So I would split them up first, then use an infinite interval from 1 to infinity and the discontinuous integrand from 0 to 1? :)

  4. lgbasallote
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    yep

  5. imagreencat
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Okay. Thanks so much for the help!!

  6. PaxPolaris
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[\large \int\limits_{0}^{\infty}{dx \over \sqrt {x} \left( x+1 \right)}\]??

  7. PaxPolaris
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    substitute \(\large \sqrt x = u\) and \(\large du =\Large \frac 12 \cdot\frac 1 {\sqrt x}\) \[\Large = 2\int\limits_0^\infty {1 \over u^2+1}du \] \[\Large = \left[ 2\tan^{-1}(u) \right]_0^\infty=2 \cdot \frac \pi 2-0 = \Huge\pi\]

  8. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy