A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • one year ago

Integrate

  • This Question is Closed
  1. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    1 Attachment
  2. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    This is an integration by parts problem, but fortunately there is a trick to reduce it a bit:

  3. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[\int\limits x^2 e^{-x/4} dx = \int\limits (4u)^2 e^{-u} d(4u) = 4^3 \int\limits u^2 e^{-u} du \]

  4. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Where I used the substitution u = x / 4 so that x = 4u and dx = d(4u) = 4 du

  5. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Now let me just call everything y instead so I can use the u in the typical form of integration by parts: \[\int\limits d(uv) = \int\limits u dv + \int\limits vdu\] Which if you drop the integral signs is merely the Liebniz rule for calculating derivatives: \[\frac{ d }{ dx } (uv) = u \frac{ d }{ dx }v + v \frac{ d }{ dx }u\] Manipulating this integral term gives: \[ \int\limits u dv = uv - \int\limits v du\] So lets take: \[4^3 \int\limits y^2 e^{-y} dy\] where again I changed the name of the original subsitution variable to y=x/4 and get to work

  6. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I used integ by parts and I got

  7. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1442185882375:dw|

  8. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Wow that was wierd the page froze....

  9. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[\int\limits x^2 e^{-x/4} dx = 4^3 \int\limits y^2 e^{-y} dy\] where we choose u=y^2 and dv = e^-y So: \[ \int\limits y^2 e^{-y} dy = -y^2 e^{-y} + 2 \int\limits y e^{-y} dy \]

  10. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Thank you so much =)

  11. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Now we choose u = y and dv = e^-y (again) which gives: \[ \int\limits y e^{-y} dy = -ye^{-y} + \int e^{-y} dy = -ye^{-y} - e^{-y} \] Putting It all together: \[ \int x^2 e^{-x/4} dx = 4^3\int\limits y^2 e^{-y} dy =4^3 ( -y^2 e^{-y} + 2 \int\limits y e^{-y} dy ) \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 4^3 ( -y^2 e^{-y} + 2 (-ye^{-y} - e^{-y})) = -64e^{-y}(y^2 -2 y + 1) \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = -64e^{-y}(y-1)^2\]

  12. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Now dont forget to substitute back in y=x/4 so your answer is in terms of the original function

  13. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[ \int x^2 e^-{x/4} dx = -64e^{-x/4}( \frac{x}{4}- 1)^2 \]

  14. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Well I messed up the exponent but you get the point

  15. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Hey how did you get that human calculator tag btw?

  16. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    The smartscore thing..

  17. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Ahhh yea I am still relatively new here, but, as you can see, I do believe I am eligible for such a title :D :D

  18. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I just need more medals

  19. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Anyways see you around

  20. 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.