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

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

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

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

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

SithsAndGiggles
 one year ago
Best 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.

ksaimouli
 one year ago
Best ResponseYou've already chosen the best response.0can u show me how to integrate sin(x^3) i mean just tell me how to do i will to the rest

SithsAndGiggles
 one year ago
Best ResponseYou've already chosen the best response.1There 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.

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

SithsAndGiggles
 one year ago
Best ResponseYou've already chosen the best response.1You 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?

ksaimouli
 one year ago
Best ResponseYou've already chosen the best response.0noo 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

SithsAndGiggles
 one year ago
Best ResponseYou've already chosen the best response.1I'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.

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

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

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

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

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

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

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

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