Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

ksaimouli

  • one year ago

integrate 3x^2e^2x dx

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

    \[\int\limits_{}^{} 3x^2e^2x dx\]

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

    i took \[u=X^2 dv=e^2x\]

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

    \[du=2x dx v=\frac{ e^2x }{ 2 }\]

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

    \[uv-\int\limits\limits_{}^{}v dv\]

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

    \[(\frac{ 3 }{ 2 }e ^{2x} x^2 -\]

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

    after that confused

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

    how to integrate v

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

    \[\int3x^2e^{2x}dx?\\ 3\int x^2e^{2x}dx\] First, integrate by parts, letting \[\begin{matrix}u=x^2& &dv=e^{2x}dx\\ du=2x\;dx& &v=\frac{1}{2}e^{2x}\end{matrix}\\ 3\int x^2e^{2x}dx=3\left[\frac{1}{2}x^2e^{2x} - \frac{1}{2}\int2xe^{2x}dx\right]\] Now, I'd let \[t=2x\\ dt=2\;dx\\ \frac{1}{2}dt=dx\] \[3\int x^2e^{2x}dx=\frac{3}{2}x^2e^{2x} - \frac{3}{2}\int te^tdt\] Integrate by parts again.

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

    thx

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

    can u show me how to integrate sin(x^3) i mean just tell me how to do i will to the rest

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

    @SithsAndGiggles

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

    There is no closed form of that integral. The best you can do is find an approximation to the function sin(x³), then integrate the approximation over a given interval.

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

    does this include series because i am seeing for the first time is this cal 2

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

    tylors series

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

    You could do that, yes. I think there's a way using the power series for that function, but I'm not so sure about that. Do you know how to find the Taylor polynomial of the function?

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

    noo i have no idea may be we will learn at the end of march next chapter but i am curious to know if u have free time u could explain or else ur wish

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

    I'll post a link to a video that explains it pretty well, if you're patient enough to watch it: https://www.khanacademy.org/math/calculus/sequences_series_approx_calc/maclaurin_taylor/v/maclauren-and-taylor-series-intuition I've only watched the older version, so I'm not sure how much better the updated one may be.

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

    ya thx i will check that out but ^ how did u get (3rd step) 2xe^2x dx

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

    That's \[v\;du = \frac{1}{2}2xe^{2x}dx,\text{ with the $\frac{1}{2}$ factored out.}\]

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

    i got that i mean it is \[\int\limits_{}^{}v \] so v=\[\frac{ 1 }{ 2 }e^2x\]

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

    i dont get where u got 2x in front of e^2x (typo ^ it is e^(2x))

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

    The 2x comes from the du term. Since u = x², we get du = 2x dx

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

    ohh so it is \[\int\limits_{}^{}v du\]

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

    Yep, integration by parts, using the u and dv substitutions, yields \[uv-\int v\;du\]

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

    okay

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

    thx

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

    You're welcome

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

    • Attachments:

Ask your own question

Ask a Question
Find more explanations on OpenStudy

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.