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ksaimouli Group Title

integrate 3x^2e^2x dx

  • one year ago
  • one year ago

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  1. ksaimouli Group Title
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    \[\int\limits_{}^{} 3x^2e^2x dx\]

    • one year ago
  2. ksaimouli Group Title
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    i took \[u=X^2 dv=e^2x\]

    • one year ago
  3. ksaimouli Group Title
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    \[du=2x dx v=\frac{ e^2x }{ 2 }\]

    • one year ago
  4. ksaimouli Group Title
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    \[uv-\int\limits\limits_{}^{}v dv\]

    • one year ago
  5. ksaimouli Group Title
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    \[(\frac{ 3 }{ 2 }e ^{2x} x^2 -\]

    • one year ago
  6. ksaimouli Group Title
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    after that confused

    • one year ago
  7. ksaimouli Group Title
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    how to integrate v

    • one year ago
  8. SithsAndGiggles Group Title
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    \[\int3x^2e^{2x}dx?\\ 3\int x^2e^{2x}dx\] 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.

    • one year ago
  9. ksaimouli Group Title
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    thx

    • one year ago
  10. ksaimouli Group Title
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    can u show me how to integrate sin(x^3) i mean just tell me how to do i will to the rest

    • one year ago
  11. ksaimouli Group Title
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    @SithsAndGiggles

    • one year ago
  12. SithsAndGiggles Group Title
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    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.

    • one year ago
  13. ksaimouli Group Title
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    does this include series because i am seeing for the first time is this cal 2

    • one year ago
  14. ksaimouli Group Title
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    tylors series

    • one year ago
  15. SithsAndGiggles Group Title
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    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?

    • one year ago
  16. ksaimouli Group Title
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    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

    • one year ago
  17. SithsAndGiggles Group Title
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    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.

    • one year ago
  18. ksaimouli Group Title
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    ya thx i will check that out but ^ how did u get (3rd step) 2xe^2x dx

    • one year ago
  19. SithsAndGiggles Group Title
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    That's \[v\;du = \frac{1}{2}2xe^{2x}dx,\text{ with the $\frac{1}{2}$ factored out.}\]

    • one year ago
  20. ksaimouli Group Title
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    i got that i mean it is \[\int\limits_{}^{}v \] so v=\[\frac{ 1 }{ 2 }e^2x\]

    • one year ago
  21. ksaimouli Group Title
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    i dont get where u got 2x in front of e^2x (typo ^ it is e^(2x))

    • one year ago
  22. SithsAndGiggles Group Title
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    The 2x comes from the du term. Since u = x², we get du = 2x dx

    • one year ago
  23. ksaimouli Group Title
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    ohh so it is \[\int\limits_{}^{}v du\]

    • one year ago
  24. SithsAndGiggles Group Title
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    Yep, integration by parts, using the u and dv substitutions, yields \[uv-\int v\;du\]

    • one year ago
  25. ksaimouli Group Title
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    okay

    • one year ago
  26. ksaimouli Group Title
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    thx

    • one year ago
  27. SithsAndGiggles Group Title
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    You're welcome

    • one year ago
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