Question

Integral of v(t)=(t^2)(e^-2t) ?

Answer 1

\[v(t)= t^2 * e ^{-2t}\]

Answer 2

I'm not sure how to u substitute or even if that's the right approach.

Answer 3

parts?

Answer 4

It looks like a by parts question if you know how to do that.

Answer 5

oh boy :/ yeah, we learned it in a hurry at the end of the year. i'll give it a shot. thank you both!

Answer 6

yeah, let us know what ya come up with :3

Answer 7

you're welcome!

Answer 8

eep got it wrong. I ended up with \[e ^{-2t}(1/3)t ^{3})-(1/12)(t^4)-e ^{-2t}\]

Answer 9

Alright. Well there are two ways Ive seen the formula given, so I'll put it the way I use and we can go from there :3

Answer 10

you rock.

Answer 11

\[u \int\limits_{}^{}v - \int\limits_{}^{}(u'\int\limits_{}^{}v)\] So the idea behind by parts is we select one part of our original function that is easy or convenient to differentiate and the remaining part to be something that is easy or convenient to differentiate. So what I notice is that e^x is always easy to differentiate or integrate, but Ill eventually get rid of x^2 if I keep differentiating it. Basically, I'm going to choose by u to be t^2 and my v is be e^(-2t). With me so far?

Answer 12

yep!

Answer 13

Alright, awesome. So now I just need to plug in some values and go from there. So I need u', but that is just 2t. I also need integral v, which is
\[-\frac{ 1 }{ 2 }e ^{-2t}\] Now we plug in.

Answer 14

\[-\frac{ 1 }{ 2 }t ^{2}e ^{-2t}-\int\limits_{}^{}(2t*-\frac{ 1 }{ 2 }e ^{-2t})\]Simplify a little and get:
\[-\frac{ 1 }{ 2 }t ^{2}e ^{-2t}+\int\limits_{}^{}te ^{-2t}\]
Now we have that integral at the end, but we still cannot just integrate it, which means we have to do by parts a second time. This time I'll have my u be t and, once again, have my v be e^(-2t).

Answer 15

where did the negative that was part of the integral go?

Answer 16

oh, out in front. gotcha.

Answer 17

Yeah, lol

Answer 18

So this time u' will just be 1 and integral v will remain the same as before. So plugging it into my by parts formula again 'I'll get:
\[-\frac{ 1 }{ 2 }t ^{2}e ^{-2t}+[-\frac{ 1 }{ 2 }te ^{-2t}-\int\limits_{}^{}(1)e ^{-2t}]\]Now we got something at the end we can integrate without needing by parts ^_^ So if I integrate that final integral and simplify, I'll get:
\[-\frac{ 1 }{ 2 }t ^{2}e ^{-2t}- \frac{ 1 }{ 2 }te ^{-2t}+\frac{ 1 }{ 2 }e ^{-2t} + C\]

Answer 19

you are an actual angel

Answer 20

Well, as long as what I did made sense xD

Answer 21

no it did, thank you so so so so much!

Answer 22

Okay, awesome ^_^ Oh, and since you're new to by parts, I thought I'd give you an 2 examples of something that might leave ya wondering what to do. If you want, lol.

Answer 23

sure! that would be great.

Answer 24

yeah since it's such a new thing i have a lot of trouble recognizing when to use it haha.

Answer 25

Right, lol. Well usually once u-sub as well as double substitution fails you're going to this. Usually, you have specifically combinations you look for that tell you by parts, but yeah. So situation number 1:
\[\int\limits_{}^{}\arcsin(x)dx\]

Answer 26

Now, this is a by parts problem, but all you see is an arcsinx. Before we had a t^2 and a e^(-2t), the parts were obvious. The parts here may not be quite as obvious.

Answer 27

What you have to do for this problem is let u = arcsin(x) and let v actually equal dx. You're allowed to choose the dx xDD So plugging all that into the by parts formula, I would have:
\[xarcsin(x) - \int\limits_{}^{}\frac{ x }{ \sqrt{1-x ^{2}} }dx\] Don't worry as much if you're not sure about derivatives of inverse functions, not a lot of people get to focus on them a lot. So now the integral I have left I can definitely do by u0substitution. I just let u = 1-x^2 and:
\[du = -2xdx ->dx = \frac{ du }{ -2x }\]Now continuing the u-substitution on the end integral I have:
\[xarcsin(x) +\frac{ 1 }{ 2 }\int\limits_{}^{}\frac{ 1 }{ \sqrt{u} }du\]After that I integrate and finish with:
\[xarcsin(x) + \sqrt{1-x ^{2}}+ C\]This is just an example of where you may need to actually choose dx to be one of your parts, which isnt always as obvious.

Answer 28

Thank you so much!! You absolutely rock, this was so helpful.

Answer 29

Mhm. Now just one more example, because this is the 2nd thing that can confuse people about by parts.

Answer 30

If you don't mind seeing, lol.

Answer 31

Not at all! That would be great if you have time but I don't want you to get annoyed or anything.

Answer 32

Nah, I'm offering because I know this stuff can trip people up xD
Okies, situation 2:
\[\int\limits_{}^{}e ^{x}\sin(x)dx\]Now our parts are obvious. In fact, it doesn't even look like it matters what I choose to be u or what I choose to be v because both are very easy to differentiate and integrate. But there is a probably that can be noticed. Before with your problem, t^2 eventually disappeared because of 2 derivatives, but e^x and sinx will never go away. You integrate OR differentiate and you'll just go in circles!

Answer 33

*problem

Answer 34

oh no why am i taking calc bc instead of stats :/

Answer 35

So now what do we do with a problem like this that goes in circles. Well, that's why I bring up this example :3 Now we usually want to choose u to be the trig function, so I'll do that. u = sinx, v = e^x. Now I plug this into my formula

Answer 36

\[e ^{x}\sin(x) - \int\limits_{}^{}e ^{x}\cos(x)dx\]Okay, so all we did was change it to cos(x). Either way, this is actually what we need to do. So I'll choose my u this time to be cos(x) and v to again be e^x

Answer 37

\[e ^{x}\sin(x) - [e ^{x}\cos(x) + \int\limits_{}^{}\sin(x)e ^{x}dx]\] Now the plus sign is there because when I took the derivative of cosx, it became negative sinx, which I factored out. Okay. awesome, back where we started. Let me distribute that negative sign in before we continue. There's a reason I keep those brackets there, VERY easy to forget and get a sign wrong. So I now get:

Answer 38

\[e ^{x}sinx -e ^{x}cosx - \int\limits_{}^{}e ^{x}\sin(x)dx\]Okay, so now here's where we magically end this chain. Let me write the FULL problem as an equation.

\[((\int\limits_{}^{}e ^{x}\sin(x)dx)) = e ^{x}\sin(x) - e ^{x}\cos(x) ((- \int\limits_{}^{}e ^{x}\sin(x)dx))\] Now notice how I surrounded the e^xsinx parts with double parenthesis. I did this to highlight them. Even though they're integrals, they're LIKE TERMS. I can add and subtract them! So I'm going to add that integral to both sides to get:

Answer 39

\[2\int\limits_{}^{}e ^{x}sinx(dx) = e ^{x}\sin(x) - e ^{x}\cos(x)\]Now I just divide both sides by two to finish.
\[\int\limits_{}^{}e ^{x}\sin(x)dx = \frac{ 1 }{ 2}(e ^{x}\sin(x) - e ^{x}\cos(x)) + C\]

Answer 40

what that's so cool.

Answer 41

Yeah, because people are not used to all of this stuff, you don't ever think that you could actually add and subtract integrals as like terms xD

Answer 42

but i feel like that is a very useful and necessary thing to know!

Answer 43

Mhm. Well, I've learned a couple tricks. Some of them are obvious, some not so obvious. Like, here's a trick. I'll give an example problem. Maybe you know the trick, maybe not :P
\[\int\limits_{}^{}\frac{ x ^{2}+2 }{ x-2 }dx\] So before I go through it, would you know what to do with that?

Answer 44

.... no :(

Answer 45

Alrighty. I just didn't wanna spam it on you if youknew what to do :P Well, mathematically, if I multiply a numerator by something I have to multiply the denominator by the same thing. Also, if I add something to one side, I either have to add it to the other side or subtract it from the same side. So what I'm going to do is add AND subtract the same thing in the numerator like this:
\[\int\limits_{}^{}\frac{ x ^{2}+2 -6 + 6 }{ x-2 }\]So I added and subtracted 6 from thetop. Well, know that we are allowed to split apart fractions as long as there is more than one term in the numerator. So watch what happens by me adding and subtracting :P

Answer 46

\[\int\limits_{}^{}\frac{ x ^{2}-4+6 }{ x-2 }dx \]this is the same as:
\[\int\limits_{}^{}\frac{ x ^{2}-4 }{ x-2 }+\int\limits_{}^{}\frac{ 6 }{ x-2 }\]Now I'll factor the left integral into (x+2)(x-2), which will allow me to cancel out the denominator to leave me with:
\[\int\limits_{}^{}x + \int\limits_{}^{}2 + \int\limits_{}^{}\frac{ 6 }{ x-2 }\] Now I just integrate and get:
\[\frac{ x ^{2} }{ 2 }+ 2x+6\ln \left| x-2 \right| + C\]

Answer 47

oh you master

Answer 48

In some rare situations, it's the only real way to do an integral. In other situations, it justspeeds up theprocess.

Answer 49

are you studying math somewhere or something? because if not, you most likely should.

Answer 50

My major is math, lol

Answer 51

good. it definitely should be haha

Answer 52

Right, lol. and I have an associates in Japanese O.o oddly enough

Answer 53

oh, cool!

Answer 54

I think so xD But yeah, I just have 5 more classes to finish my associates in math. Still got 4 more math classes. I could only do 2 this semester because of schedule conflict D:

Answer 55

oh no! hopefully next semester you'll be able to fit more in. anyways, thanks a bajillion for your help, i really do appreciate it. have a great night :)

Answer 56

Night :3

Answer 57

\[\int\limits \ln x \text{ } dx\]this is one more you should see...
\[\text{Let }u=\ln x,\text{ } dv = dx \text{ so... } du = \frac{ dx }{x }, v=x. \text{ Then }\]
\[\int\limits \ln x \text{ }dx = x \ln x -\int\limits x \frac{ dx }{ x }=x \ln x -\int\limits dx=x \ln x - x\]