Question

Definite Integral
Goes like this....

Answer 1

\[\int\limits_{2}^{?0} t^2 * e^t\]

Answer 2

oops, the upper limit is 2 and the lower limit is 0, i got it mixed up, sry.

Answer 3

@OOOPS

Answer 4

So what do you think ?

Answer 5

How should we approach the problem ?

Answer 6

well I got it to [t^3/3*e^t] and then just put in the upper and lower limits which put I got the wrong answer.

Answer 7

what am I doing wrong?

Answer 8

You forgot that you have to integrate the *product* of two functions.

Answer 9

oh so I should do u'v+u'v then integrate them?

Answer 10

No no that is for differentiation. For integration you would have to use integration by parts.

Answer 11

that is product rule , not in integration

Answer 12

\[
\int t^2 * e^tdt = \int t^2 * d(e^t) \\
\int u * dv = uv - \int vdu
\]

Answer 13

Well @Xlegalize is really not wrong since integration by parts is the product rule: \[\Large (uv)'=u'v+uv'\]\[\Large \int\limits (uv)'dx=\int\limits u'v dx+\int\limits uv'dx\]\[\Large uv=\int\limits \frac{du}{dx}v dx+\int\limits u \frac{dv}{dx}dx\]\[\Large uv=\int\limits vdu+\int\limits u dv\]\[\Large uv-\int\limits vdu=\int\limits u dv\]

Answer 14

so why do I get the wrong answer then?

Answer 15

it says the answer is 12,778 But I get it around 18.

Answer 16

Show us your steps, it's the only way we can correct your mistakes.

Answer 17

okey so the integrating t^2 is t^3/3 and integrating e^t is e^t and putting in the limits which is 2 and 0 u get [2^3/3 * e^2] - [0^3/3*e^0]

Answer 18

Ahh, what you must do is integrate one and differentiate the other, not integrate both separately.

Answer 19

Could you draw equation of it? im not quite sure..

Answer 20

Ok, so for example, let's do a similar example to show you how it works. =)

Answer 21

okey:)

Answer 22

Okey wait, I think I got it, give me a few and I'll come back to you:)

Answer 23

\[\Large \int\limits udv = uv-\int\limits v du\] This is our formula that we're going to use to convert the integral on the left into the integral with the uv on the right to hopefully make it simpler for us. What we must determine is what is u and what is dv?

Generally what you will pick for u is something that goes away after you take the derivative. So, like x will become just a constant after you take its derivative while e^x won't go away. This should make more sense by the time we're through if it's still a little uncertain right now. Ok so this will be my example:
\[\Large \int\limits xe^xdx\]
Now I need to choose a u and a dv. We'll choose\[\Large x=u \ \ \ \ e^xdx = dv\] Now we take the derivative of the equation of u and integrate the one on the right \[\Large dx=du \ \ \ \ e^x = v\] And we plug in.

I'll stop the example here then and wait for you. =)

Answer 24

so [2te^t] and then we plug in the limits give:
2(2e^2)-2e^0 = 4e^2-2 = 14,77-2= 12,77 :):)

Answer 25

I haven't actually solved the problem myself yet, but as long as you did integration by parts twice you probably did it correctly. =)

Answer 26

yes I did, but I'll tell you this integration section, its too wide, by that I mean there are so many ways to it, its impossible to keep them apart, if you know what I mean. :)

Answer 27

Yeah, I remember when I took it I felt very rushed like it was very fast paced. But if you can find a common thread between things or kind of get a beat on what it is, things will make more sense.

A good tool that helped me get integration by parts straight before I had a good intuition was to use:

detail

which basically means that the "d" stands for the "dv" part that you should pick should be picked in a certain priority following these letters.

e - e^x
t - trig
a - agebraic (x^3, etc)
i - inverse trig (arcsin(x))
l - log

So if you have to integrate something like: \[\Large \int\limits \sin^{-1}(x)\ln(x)dx\] Then you should pick \[\Large \ln(x)=u \ \ \ \ \sin^{-1}(x)dx = dv\]

That might seem like a hard integral, it might even be unsolvable idk, but this is one of the best strategies I've found.

Answer 28

okey thanks, I'll try to get this intution going, because its only getting harder:)

Answer 29

Yeah, I think if you just allow yourself to become interested in the subject and try to do some weird or interesting things, the practice you get will help build your intuition. And wolframalpha.com is probably your best friend for solving integrals to give you something to check.

Some ideas:

What if you try to do \[\Large \int\limits x^2dx\] as integration by parts by separating it apart? \[\Large \int\limits x*xdx\] \[\Large x = u \ \ \ \ \ xdx = dv\]

Alright, I'm out of here, good luck haha