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ksaimouli

  • 3 years ago

integrate 3x^2e^2x dx

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  1. ksaimouli
    • 3 years ago
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    \[\int\limits_{}^{} 3x^2e^2x dx\]

  2. ksaimouli
    • 3 years ago
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    i took \[u=X^2 dv=e^2x\]

  3. ksaimouli
    • 3 years ago
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    \[du=2x dx v=\frac{ e^2x }{ 2 }\]

  4. ksaimouli
    • 3 years ago
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    \[uv-\int\limits\limits_{}^{}v dv\]

  5. ksaimouli
    • 3 years ago
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    \[(\frac{ 3 }{ 2 }e ^{2x} x^2 -\]

  6. ksaimouli
    • 3 years ago
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    after that confused

  7. ksaimouli
    • 3 years ago
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    how to integrate v

  8. SithsAndGiggles
    • 3 years ago
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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.

  9. ksaimouli
    • 3 years ago
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    thx

  10. ksaimouli
    • 3 years ago
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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

  11. ksaimouli
    • 3 years ago
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    @SithsAndGiggles

  12. SithsAndGiggles
    • 3 years ago
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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.

  13. ksaimouli
    • 3 years ago
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    does this include series because i am seeing for the first time is this cal 2

  14. ksaimouli
    • 3 years ago
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    tylors series

  15. SithsAndGiggles
    • 3 years ago
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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?

  16. ksaimouli
    • 3 years ago
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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

  17. SithsAndGiggles
    • 3 years ago
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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.

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

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

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

  21. ksaimouli
    • 3 years ago
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    i dont get where u got 2x in front of e^2x (typo ^ it is e^(2x))

  22. SithsAndGiggles
    • 3 years ago
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    The 2x comes from the du term. Since u = x², we get du = 2x dx

  23. ksaimouli
    • 3 years ago
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    ohh so it is \[\int\limits_{}^{}v du\]

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

  25. ksaimouli
    • 3 years ago
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    okay

  26. ksaimouli
    • 3 years ago
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    thx

  27. SithsAndGiggles
    • 3 years ago
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    You're welcome

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