Question

integrate 3x^2e^2x dx

Answer 1

\[\int\limits_{}^{} 3x^2e^2x dx\]

Answer 2

i took \[u=X^2 dv=e^2x\]

Answer 3

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

Answer 4

\[uv-\int\limits\limits_{}^{}v dv\]

Answer 5

\[(\frac{ 3 }{ 2 }e ^{2x} x^2 -\]

Answer 6

after that confused

Answer 7

how to integrate v

Answer 8

\[\int3x^2e^{2x}dx?\\
3\int x^2e^{2x}dx\]
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.

Answer 9

thx

Answer 10

can u show me how to integrate sin(x^3) i mean just tell me how to do i will to the rest

Answer 11

@SithsAndGiggles

Answer 12

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.

Answer 13

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

Answer 14

tylors series

Answer 15

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?

Answer 16

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

Answer 17

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.

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

okay

Answer 26

thx

Answer 27

You're welcome