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

1. ksaimouli

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

2. ksaimouli

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

3. ksaimouli

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

4. ksaimouli

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

5. ksaimouli

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

6. ksaimouli

after that confused

7. ksaimouli

how to integrate v

8. SithsAndGiggles

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

thx

10. ksaimouli

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

@SithsAndGiggles

12. SithsAndGiggles

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

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

14. ksaimouli

tylors series

15. SithsAndGiggles

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

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

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

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

19. SithsAndGiggles

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

20. ksaimouli

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

21. ksaimouli

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

22. SithsAndGiggles

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

23. ksaimouli

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

24. SithsAndGiggles

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

25. ksaimouli

okay

26. ksaimouli

thx

27. SithsAndGiggles

You're welcome