Question

Integral 4xe^-x

Answer 1

you will need to use integration by parts. There may be a simpler method I'm not seeing, however.

Answer 2

\[udv = uv-\int\limits_{}^{}duv\]

Answer 3

set 4x as your u and e^-x as your dv

Answer 4

I think I'm even more confused now. I don't use methods like that in my maths, and I cant find my notes on integration by parts

Answer 5

Are you in calc 1 or calc 2

Answer 6

see this http://mathnow.wordpress.com/2009/10/14/liate-ilate-and-detail/

Answer 7

as @VeritasVosLiberabit said you should use integration by parts and the trick in it is to choose u and v , in the previous link there's an acronym to remember which one to pick as u and v. hope its helpful

Answer 8

@SandeepReddy I wonder if there is another way (calc 1 method?) that I am forgetting.

Answer 9

to my knowledge i can assure that this is the only way to do it

Answer 10

Just hit me that you guys are probably american? I'm studying in Scotland and its a uni module I'm doing thats got the integration in it, so the method I've been shown may differ.. I can find notes on integration by substitution, is that the same thing? Tbh I'm totally unsure of integration as a whole and dont even really know how to integrate any of the parts at all. I'm just so lost!

Answer 11

@adrianne92 So if this is your first calc class, that is what makes me think there may a clever trick to figuring this out. But the only way I can think to solve this at the moment is using integration by parts

Answer 12

I've always seen this type of integral being solved with integration by parts.

Answer 13

is integration by parts similar to the function of a function rule in differentiation?

Answer 14

I think this page will explain it well:
http://www.math.wisc.edu/~park/Fall2011/integration/Integration%20by%20parts.pdf

It also gives a proof for it

Answer 15

you have integration by substitution right?, so in that method, you will first find an antiderivative ( integral ) of the function and then substitute the end values and substract them right?

Answer 16

yes thats right, is this the same thing then?

Answer 17

Integration by parts is not the same as the "regular" substitution rule. Please check here it will explain it in more detail, and here are loads of examples:
http://www.math.wisc.edu/~park/Fall2011/integration/Integration%20by%20parts.pdf

Answer 18

there*
Including an example similar to yours

Answer 19

@adrianne92 , here integration by parts is just finding anti derivative of the function given to you,
and integration by substitution needs you to find the anti derivative ( as in integration by parts) and then you will substitute the end values given to you

Answer 20

so you just need to find the anti derivative , do you get it?

Answer 21

I think so, I'm having a look at that link @kirbykirby gave me just now and I think your explanation there will help me with using my substitution notes just as a guideline