Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

integrate 3x^2e^2x dx

I got my questions answered at in under 10 minutes. Go to now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly


Get your free account and access expert answers to this and thousands of other questions

\[\int\limits_{}^{} 3x^2e^2x dx\]
i took \[u=X^2 dv=e^2x\]
\[du=2x dx v=\frac{ e^2x }{ 2 }\]

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

\[uv-\int\limits\limits_{}^{}v dv\]
\[(\frac{ 3 }{ 2 }e ^{2x} x^2 -\]
after that confused
how to integrate v
\[\int3x^2e^{2x}dx?\\ 3\int x^2e^{2x}dx\] 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.
can u show me how to integrate sin(x^3) i mean just tell me how to do i will to the rest
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.
does this include series because i am seeing for the first time is this cal 2
tylors series
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?
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
I'll post a link to a video that explains it pretty well, if you're patient enough to watch it: I've only watched the older version, so I'm not sure how much better the updated one may be.
ya thx i will check that out but ^ how did u get (3rd step) 2xe^2x dx
That's \[v\;du = \frac{1}{2}2xe^{2x}dx,\text{ with the $\frac{1}{2}$ factored out.}\]
i got that i mean it is \[\int\limits_{}^{}v \] so v=\[\frac{ 1 }{ 2 }e^2x\]
i dont get where u got 2x in front of e^2x (typo ^ it is e^(2x))
The 2x comes from the du term. Since u = x², we get du = 2x dx
ohh so it is \[\int\limits_{}^{}v du\]
Yep, integration by parts, using the u and dv substitutions, yields \[uv-\int v\;du\]
You're welcome

Not the answer you are looking for?

Search for more explanations.

Ask your own question