## ksaimouli Group Title integrate 3x^2e^2x dx one year ago one year ago

1. ksaimouli Group Title

$\int\limits_{}^{} 3x^2e^2x dx$

2. ksaimouli Group Title

i took $u=X^2 dv=e^2x$

3. ksaimouli Group Title

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

4. ksaimouli Group Title

$uv-\int\limits\limits_{}^{}v dv$

5. ksaimouli Group Title

$(\frac{ 3 }{ 2 }e ^{2x} x^2 -$

6. ksaimouli Group Title

after that confused

7. ksaimouli Group Title

how to integrate v

8. SithsAndGiggles Group Title

$\int3x^2e^{2x}dx?\\ 3\int x^2e^{2x}dx$ 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 Group Title

thx

10. ksaimouli Group Title

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 Group Title

@SithsAndGiggles

12. SithsAndGiggles Group Title

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 Group Title

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

14. ksaimouli Group Title

tylors series

15. SithsAndGiggles Group Title

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 Group Title

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 Group Title

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 Group Title

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

19. SithsAndGiggles Group Title

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

20. ksaimouli Group Title

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

21. ksaimouli Group Title

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

22. SithsAndGiggles Group Title

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

23. ksaimouli Group Title

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

24. SithsAndGiggles Group Title

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

25. ksaimouli Group Title

okay

26. ksaimouli Group Title

thx

27. SithsAndGiggles Group Title

You're welcome