Question

find the indefinite integral

Answer 1

\[\int\limits \frac{4x^2}{e^x} dx\]

Answer 2

parts again right?

Answer 3

\[\int\limits 4x^2 * e^{1/2}dx\]

Answer 4

I would assume so...

Answer 5

whoa hold on

Answer 6

\[\int 4x^2e^{-x}dx\]

Answer 7

oh woops... I forgot that's a sqrt() lol this major headache is killing me...

Answer 8

u = 4x^2
1/8 du = xdx

dv = e^-xdx
v = -e^-x

Answer 9

ok then it should be good from here
you are going to need parts twice to reduce the power of \(x^2\) and the bookkeeping is going to be annoying because of the \(-x\) in the exponent. but that is what you have to do

Answer 10

this can't be any harder than the last question which was ridiculous...

Answer 11

and for pete's sake bring the 4 out front so you don't have to keep messing with it

Answer 12

in fact ignore the 4 entirely, then stick a 4 in front of your answer at the end

Answer 13

ok

Answer 14

maybe :P

Answer 15

What satellite73 said.
I get:
\[-4e^{-x}(x^2+2x+2)\]
I will double check.

Answer 16

wut.............

Answer 17

amistre has a really snap way of doing these using a little chart, but i forget it
he can do parts a bunch of times in a breeze
youtube it and i bet you will find it

Answer 18

so \[x^2 * -e^{-x} + \int\limits e^{-x} xdx\]

Answer 19

you're missing an x there, you have only decreased the exponent by one, so it should be +2 int xe^-x

Answer 20

besides that, you're on the right track.

Answer 21

hmm?

Answer 22

\[-x^2-e^{-x} + 2 \int\limits_{}^{}xe^{-x}\]
for the beginning

Answer 23

ohh oops, there is one - too many, what you wrote is correct indeed.

Answer 24

I knew I forgot a 1/2

Answer 25

yes and the x be next to the integral

Answer 26

mhmhm

Answer 27

\[x^2 * -e^{-x} +\frac{1}{2} x * -e^{-x} + \int\limits -e^-xdx\]

Answer 28

\[x^2 * -e^{-x} +\frac{1}{2} x * -e^{-x} + e^{-x} +c\]

Answer 29

I meant just dx

Answer 30

where does this 1/2 come from again? I am not seeing that to be honest.

Answer 31

nvm should be a 2 I was subbing in for 1/2 du = dx.

Answer 32

Man I wish I didnt; have this horrible headache of doom.

Answer 33

good, so it should be a two in the last two terms, then you got it.

Answer 34

last 2?

Answer 35

u = x du = dx

Answer 36

\[4\int x^2 e^{-x} = -x^2e^{-x}+2\int xe^{-x}\]
so
\[2\int x e^{-x} = -xe^{-x} + \int e^{-x} = -2xe^{-x} -2e^{-x}\]

Answer 37

okies.

Answer 38

\[-4e^{-x}(x^2+2x+2)\]

Answer 39

so it distributes over?

Answer 40

yes

Answer 41

the times 4 can be ignored because it's in front of the first integral, this means that whatever the solution will be, just multiply that times 4 and you have got taken care of that scalar quantity again.

Same goes for the 2nd integral, at the end just remember to multiply everything times two (after that integral)

Answer 42

I just wasn't sure it was distributed to all of the parts of the integral...

Answer 43

onto the next question! http://openstudy.com/study#/updates/4ffb5a30e4b00c7a70c463f4