Question

How do you evaluate the integral 3/2xe^x dx

Answer 1

Do you know how to do integration by parts?

Answer 2

Well I learned that in Cal 2 but don't remember lol..

Answer 3

Yeah, that would be where ya learn it, haha. Well, there are two formulas that I know of for it, but this is the one I use:

\[u \int\limits_{}^{}v - (\int\limits_{}^{}u'\int\limits_{}^{}v)\]

Answer 4

Weird to type that on here, I swear.

Answer 5

But okay, so we need to choose a u and we need to choose a v. We choose our u based on what is convenient to take the derivative of and we choose our v based on what is convenient to take an integral of. Kinda recall some of that?

Answer 6

Integration using parts

Answer 7

Yes its coming back sort of lol

Answer 8

Lol, alright, cool. So if we had to pick a u (easy or convenient to differentiate) and a v (easy or convenient to integrate), what would we choose? :P

Answer 9

I would choose 3/2x to be u and e^x to be v

Answer 10

You don't need to quite do that. Remember, we can factor out that 3/2 and just multiply it in later. Might be easier to do.

Answer 11

Also how do you know when to use integration by parts?

Answer 12

so x would be u and e^x would be v

Answer 13

Usually you have something like xe^x, e^x(sinx), sin(x)ln(x), there are usually terms that do not normally match up in terms of normal algebra. Of course you can also know because u-substitution, x-substitution, or any of that won't work.
And be careful, you can't just choose x or e^x, those are in the denominator, so they must be (1/x) = u and (1/e^x) = v. Or if you prefer, x^-1 = u, e^(-x) = v

Answer 14

Sorry I wrote the question wrong -.-. Its actually 3/2 * xe^x dx . The xe^x are not in the deniminator

Answer 15

Oh!. Okay, that makes this integral wayyyy easier. (nightmare if its in denom). Okay, so yes, u = x and v = e^x. So now v itself isn't really needed, we just need integral v.....which is e^x, haha. So if you got the formula, try plugging in the values and see what ya get. Does the formula make sense?

Answer 16

Ill try it right now

Answer 17

kk, ill do it, too.

Answer 18

wait wouldn't e^x be the dv?

Answer 19

and v ends up being e^x still

Answer 20

Yes, it does stay as e^x after integrating it. That's why we picked v to be e^x. it's integral is easy to do. It doesn;t matter that it's the same exact thing, you still plugit into theformula.

Answer 21

So I got 3/2 * xe^x - 3/2e^x

Answer 22

+ C xDD

Answer 23

oh yeah that C...Lol thanks again :D

Answer 24

Haha, yeah, I hate the damn thing, too -_- But yep, np ^_^

Answer 25

I really should know these things but I forgot a lot of stuff from Cal 2 and im paying for it now in Cal 3 -.-

Answer 26

The only reason I know it is because I'm a math tutor at my college. Itforces me to review a lot. I'm taking discrete math and diff eqs here in a couple weeks. I'd be in calc 3, too, but schedule conflict x_x

Answer 27

ahhh lol. I took discrete math already I believe. I don't have to take diff eqs thank god. I might have to tutor or something so I can remember this stuff..

Answer 28

How was discrete? It looks like a math logic class, haha. But yeah, it helps me remember for sure. I just had to get a recommendation, take a test, then do an interview. I swear, you get enough people coming in for calculus and you never forget it xD

Answer 29

Haha nice. Discrete wasn't too bad. yes its basically logic and proofs. Are you a CS major? I had to take discrete only because I was a CS major but when I switched to math it wasn't a requirement.

Answer 30

Nah, im actually a pure math major. I still have to take the class, though. I have that class, diff eqs, and calc 3 left before 300-400 level stuff. I do have to do some CSstuff, though, I'm taking a CS class in the fall as well.

Answer 31

haha have fun with CS. I like computers and all but I couldn't program for pellet...

Answer 32

wow pellet lol

Answer 33

Yeah, seems like thats exactly what im in for :/ Textbook itself is C++ programming.

Answer 34

ah I did Java while I was in CS until they changed the entire thing to start in C++

Answer 35

Its really not my thing, but looks like I have to do a few things that could be related to CS. Forgot, I also have to take matrix algebra x_x

Answer 36

Ah Im guessing Matrix Algebra is basically Linear Algebra? I just took that last month in a summer semester. You could imagine everything was compressed into a month >.< lol it's not too bad. I prefer linear algebra over calculus 3!

Answer 37

Oh wow, lol. Yeah, there were no math classes I could take in the summer or I would have. And yeah, pretty much is. I eventually have to do linear algebra, too. Im going to try and do that in the same semester as calc 3. Is calc 3 really that bad or is just linear algebra more fun? xD

Answer 38

both..lol I guess you could say Calculus 3 is harder if you don't remember stuff from the previous cal's.. But calculus 2 was definitely worst than calculus 3.

Answer 39

you'll do fine in cal 3 though because you've been helping me with calculus 3 stuff..lol

Answer 40

I think I just lucked out for calc 2. We were allowed formula sheets, made the class very easy. Now when I took calc 1, guy couldnt teach worth a damn, I was basically self-taught in that class. And yeah, someof the calc 3 stuff isnt bad xD Ive perused through the topics and picked up on some of the ones I thought were simpler. So i've seen the partial derivatives and double/triple integrals. Not much more than that, though.

Answer 41

Nice. Im basically teaching myself for calculus 3 right now lol. Yeah partial derivatives isn't bad. We just covered triple integrals a couple of days ago. Still doing the hw for double integrals -.- So much hw....

Answer 42

Not good :/ How far into the class are you? Like, how much more do you have to worry about learning?

Answer 43

I think about 2 weeks left O.O.. thats the scary thing lol. I have my 2nd midterm on Monday and then I think the final is on Friday..

Answer 44

This is also a 1 month summer course

Answer 45

Ah. Haha, speed calc 3 course, geez O.o Sounds bad. Well, the only reason I ever found this site was because of this paul's online notes thing. Lot of people have seen it, lot of peoplel haven't. But anything I've seen from calc 3 I studied there. You know of the site?

Answer 46

Nope I never heard of it but I just looked it up. Maybe it'll come in handy though :p.

Answer 47

Yeah, professor put his whole class notes on this open website, so that's pretty awesome. Might help out. And if you come up with another question maybe I can refer to it and help out, too, lol.

Answer 48

haha sweet. yeah ill check it out

Answer 49

Mhm. I'll look through it some, too, I need to learn it anyway xD I'm looking at something that might be upcoming. Double integrals in polar coordinates.

Answer 50

yep we did that but I haven't taught myself yet lol

Answer 51

Ah, already did it xD Yeah, im just looking through it now. I was wondering when Id see that again. Hasnt been since trig really that I did much with that.

Answer 52

lol yeah you will deal with polar coordinates in cal 3 >.<

Answer 53

Yeah, just going through it now. This first example problem doesn't look too bad. Just the whole inequality thing is a new concept.

Answer 54

for the integral of (x*y*e^xy^(2)) how would I tackle that? Using IBP again?

Answer 55

with respect to y

Answer 56

\[\int\limits_{}^{}xye ^{xy ^{2}}dy\] Is that correct? You only need it in respect to y?

Answer 57

correct

Answer 58

Then yeah, we just pretend that x is merely a constant and go from there. In this case it comes off nicely. Just let u = xy^2 and pretty much everything works out.

Answer 59

Come up with something?

Answer 60

ya and its totally off lol.. I left u = xy^2, v= y^2/2, du = 2xy, dv = y
Plugging into IBP formula: uv - integral(vdu) => xy^2 * y^2/2 - integral(y^2/2*y) dy

Answer 61

* let not left

Answer 62

Lol, we dont need IBP, just treat it like a normal integral o.o

Answer 63

I meant u as in u-substitution o.o

Answer 64

rofl oops >.< brb

Answer 65

Lol, alright.

Answer 66

I'm gonna have to head out, so I'll just show ya most of it before I go:

\[\int\limits_{}^{}xye ^{xy ^{2}}dy\]
let u = xy^2, meaning du = 2xydy and dy = (du/2xy)

\[\int\limits_{}^{}xye ^{u}\frac{ du }{ 2xy }\]

From there you can kinda see how it'll go. I just pretended x was a constant and started integrating in respect to y. Should be able to finish there. Later.

Answer 67

@psymon I see that part you did above but what now that you have u in the integral?

Answer 68

Well, since I called that portion of the integral u, I keep it as u until after I integrate. Basically, you treat it as if it were a simple x up there. As a formula for the integration of e^x, its actually listed as e^u. So if you got everything up to that point out of the way, you would integrate e^u, which is just e^u, multiply the 1/2 back in and replace u with the original xy^2. If that makes any sense.

Answer 69

ok got it thx

Answer 70

Okay, cool xD glad it made some sense then.