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ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
\[\int\limits_{}^{} 3x^2e^2x dx\]
 one year ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
i took \[u=X^2 dv=e^2x\]
 one year ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
\[du=2x dx v=\frac{ e^2x }{ 2 }\]
 one year ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
\[uv\int\limits\limits_{}^{}v dv\]
 one year ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
\[(\frac{ 3 }{ 2 }e ^{2x} x^2 \]
 one year ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
after that confused
 one year ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
how to integrate v
 one year ago

SithsAndGiggles Group TitleBest ResponseYou've already chosen the best response.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.
 one year ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
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

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
@SithsAndGiggles
 one year ago

SithsAndGiggles Group TitleBest ResponseYou've already chosen the best response.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.
 one year ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
does this include series because i am seeing for the first time is this cal 2
 one year ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
tylors series
 one year ago

SithsAndGiggles Group TitleBest ResponseYou've already chosen the best response.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?
 one year ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.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
 one year ago

SithsAndGiggles Group TitleBest ResponseYou've already chosen the best response.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/maclaurenandtaylorseriesintuition I've only watched the older version, so I'm not sure how much better the updated one may be.
 one year ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
ya thx i will check that out but ^ how did u get (3rd step) 2xe^2x dx
 one year ago

SithsAndGiggles Group TitleBest ResponseYou've already chosen the best response.1
That's \[v\;du = \frac{1}{2}2xe^{2x}dx,\text{ with the $\frac{1}{2}$ factored out.}\]
 one year ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
i got that i mean it is \[\int\limits_{}^{}v \] so v=\[\frac{ 1 }{ 2 }e^2x\]
 one year ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
i dont get where u got 2x in front of e^2x (typo ^ it is e^(2x))
 one year ago

SithsAndGiggles Group TitleBest ResponseYou've already chosen the best response.1
The 2x comes from the du term. Since u = x², we get du = 2x dx
 one year ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.0
ohh so it is \[\int\limits_{}^{}v du\]
 one year ago

SithsAndGiggles Group TitleBest ResponseYou've already chosen the best response.1
Yep, integration by parts, using the u and dv substitutions, yields \[uv\int v\;du\]
 one year ago

SithsAndGiggles Group TitleBest ResponseYou've already chosen the best response.1
You're welcome
 one year ago
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