Question

integration of xe^(-x/2)

Answer 1

\(\large\color{black}{ \int\limits_{ }^{ }\ xe^{-x\div2}~dx }\) this?

Answer 2

You can do this sort of trick with the product rule: \[\Large (xe^x)' = (x)'e^x+x(e^x)' \\ \Large (xe^x)'=e^x+xe^x \\ \Large \int\limits (xe^x)' dx = \int\limits e^x dx + \int\limits xe^x dx \\ \Large xe^x- e^x= \int\limits xe^x dx \]

Answer 3

wait you just proved integration by parts. I saw this in khan academy once.

Answer 4

I mean I am just giving the kid friendly version of integration by parts. It is exactly the product rule just from a slightly different perspective.

Answer 5

well, with e^x it is certainly easier, because it's integral and derivative on it itself are all the same:)

Answer 6

\[\LARGE \int\limits_{ }^{ }xe^{-\frac{1}{2}x}~dx\]
\(\large\color{black}{ u=-\frac{1}{2}x,~~~~~~x=-2u ,~~~~~~dx=-2~du }\)

\[(-2)(-2)\LARGE \int\limits_{ }^{ }ue^{u}~du\]

\[4\LARGE \int\limits_{ }^{ }ue^{u}~du\]

now, it should look a bit better, do you integration By Parts.

Answer 7

differentiate the "u" to reduce the u.

Answer 8

differentiate the "u" to reduce the u.

Answer 9

I was about to put up a reply, but disconnected, I'll just draw it...