A community for students. Sign up today!
Here's the question you clicked on:
 0 viewing

This Question is Closed

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

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

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

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

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

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

SithsAndGiggles
 2 years ago
Best ResponseYou've already chosen the best response.1Yep, integration by parts, using the u and dv substitutions, yields \[uv\int v\;du\]
Ask your own question
Ask a QuestionFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.