Question

how do you integrate xe^(2x)dx?

Answer 1

integral from 0 to 2

Answer 2

Use LATE principle?

Answer 3

oh im using integration by parts and i know after we find the simplest from we plus in the limits into x but im not sue why the answer is 1/4 (3e^4+1)
where did they get the 1/4 th

Answer 4

-1/4+x^2/2+2x^3/3+x^4/2+0(x^5)

Answer 5

\[\Large \int\limits \color{royalblue}{x} e^{2x}\;dx\]So by parts huh?
Ok so we want the blue x to disappear right? So we'll let that be our `u`.
Do you understand how to integrate the e^(2x) to find your `v`?

Answer 6

\[\Large \int\limits\limits \color{royalblue}{x} e^{2x}\;dx\quad=\quad \frac{1}{2}x e^{2x}-\frac{1}{2}\int\limits e^{2x}\;dx\]By parts gives us this, yes? :o

Answer 7

yes ive gotten that so far

Answer 8

plugging in the limits is the confusing part.

Answer 9

Sorry was cooking din din D:

Answer 10

So finishing up the integration gives us:\[\Large \frac{1}{2}x e^{2x}-\frac{1}{4}e^{2x}\]Factoring some stuff:\[\Large \frac{1}{4}e^{2x}\left(2x-1\right)\]
And now we need to plug in some limits? Hmm

Answer 11

So at the upper limit we get:\[\Large \frac{1}{4}e^{4}(4-1)\]And then at the lower limit:\[\Large \frac{1}{4}e^0(-1)\]

Answer 12

Giving us a grand total offffff:\[\Large \frac{1}{4}e^{4}(4-1)\quad - \quad \left[\frac{1}{4}e^0(-1)\right]\]

Answer 13

The step where I factored out all of the stuff:
1/4 and e^2x, was that confusing at all?
Do you understand why we were left with 2x in the first term?

Answer 14

the factoring was indeed confusing but if i stare at theses steps for a while im sure ill get the hang of it. :) thank you very much! its a grand help!

Answer 15

Try to keep this little math tid bit in mind:\[\Large 2+\frac{1}{4}\]If we want to factor 1/4 out of each term, we're `dividing` each term by 1/4.
Which means we're actually `multiplying` each term by 4 as we factor our 1/4 out, right? :)
\[\Large \frac{1}{4}\left(8+1\right)\]

Answer 16

That's just an example, not numbers from our actual problem.
A little tricky ^^ just something to keep in mind.

Answer 17

ah, i am starting to connect things. thank you again. its been a great help.