A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

AmTran_Bus

  • one year ago

Can someone help me with this integration by parts?

  • This Question is Closed
  1. AmTran_Bus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1433873040803:dw|

  2. AmTran_Bus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1433873069974:dw|

  3. AmTran_Bus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    That is supposed to say du= 2r dr

  4. dumbcow
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    you forgot the sqrt around the "4+r^2" actually i dont think you need int by parts here .... just the substitution --> u = 4+r^2 du = 2r dr

  5. AmTran_Bus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Hum. This problem is from the integration by parts chapter in the Stewart book :(

  6. AmTran_Bus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    But ok. Lets do it your way! Can you guide me through it?

  7. dumbcow
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    sure ... double check you copied the question correctly though

  8. AmTran_Bus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Ok. I did.

  9. dumbcow
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    after substitution you get \[\frac{(u-4) r}{\sqrt{u}} * \frac{du}{2r}\]

  10. dumbcow
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    the "r" cancels leaving \[\frac{1}{2} \int\limits \frac{u-4}{\sqrt{u}} du\]

  11. AmTran_Bus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Ok, can you explain the after substitution step? Is u=4+r^2?

  12. dumbcow
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    yes so since u = 4 +r^2 , if you solve for r^2 you get r^2 = u-4 --> r^3 = r^2 *r = (u-4)r

  13. AmTran_Bus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Got ya. Thanks.

  14. AmTran_Bus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Don't we have to solve for out limits?

  15. AmTran_Bus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    *our

  16. dumbcow
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    yes but first you must find integral .... then plug in limits at the end

  17. AmTran_Bus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

  18. dumbcow
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    can you finish the integral? \[\frac{u-4}{\sqrt{u}} = u^{1/2} -4u^{-1/2}\]

  19. phi
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    FYI if you want to use integration by parts, notice that to integrate \[ (4+r^2)^{-\frac{1}{2} } \] Taking that as a hint, write the problem as \[ \int \frac{r^3}{\sqrt{4+r^2}} dr = \int r^2 (4+r^2)\ r\ dr \] and let \[ u= r^2 \\ dv = (4+r^2)\ r\ dr \]

  20. AmTran_Bus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Ok. So the (4+r^2)^(-1/2) was basically put on top now, right?

  21. phi
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    yes, only because I don't like square roots.

  22. phi
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    to integrate \[ (4+r^2)^{-\frac{1}{2} } \] we need 2 r dr

  23. AmTran_Bus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    So we havae|dw:1433877516461:dw|

  24. AmTran_Bus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Im just trying to get to what you did to be sure I get it.

  25. phi
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    so split the r^3 into r^2 and r (and put the r next to the dr)

  26. AmTran_Bus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Ok. r^2 *r

  27. phi
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    yes. so it should look like this \[ \int \frac{r^3}{\sqrt{4+r^2}} dr = \int r^2 (4+r^2)^{-\frac{1}{2}}\ r\ dr\]

  28. AmTran_Bus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Very nice. So now I can just do the regular U and dv stuff.

  29. phi
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    yes, u= r^2 dv = what is left over

  30. AmTran_Bus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Thank you so very much. I still have just a small issue with how you got|dw:1433877820455:dw|

  31. AmTran_Bus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Sorry for that mess. And for asking you so much.

  32. AmTran_Bus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I get how you brought the sqrt up top.

  33. AmTran_Bus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1433877903541:dw|

  34. phi
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    you don't have to change the square root to an exponent, but I find it easier to think about|dw:1433877934816:dw|

  35. phi
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \[ \frac{1}{x}= x^{-1} \] \[ \sqrt{x} = x^\frac{1}{2} \] \[ \frac{1}{\sqrt{x}}=\frac{1}{x^\frac{1}{2}}=x^{-\frac{1}{2}} \]

  36. AmTran_Bus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Thanks so much phi. I am really glad you showed me how to do it this way.

  37. phi
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    You still have to do the integral

  38. AmTran_Bus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I know, from all the U= and DV= stuff. But I'm pretty good with that.

  39. 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

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.