Question

use integration by parts twice to find the exact value of.... see below

Answer 1

\[\int\limits_{0}^{4}x ^{2}e ^{\frac{-x }{ 4 }}\] dx

Answer 2

(the dx should be on the same line)

Answer 3

Let
u=x^2 dv=e^{-x/4}dx
du=2xdx v=-4e^(-x/4)

Answer 4

\[-4x^2e^{-x/4}+4\int\limits e^{-x/4}(2x)dx\]

Answer 5

Do the same integration by parts

Answer 6

yep

Answer 7

u=2x
du/dx = 2

Answer 8

and dv/dx = -4e^(-x/4)
v = 16e^(-x/4) ????

Answer 9

which would become

Answer 10

oh no actually,

Answer 11

dv/dv = e−x/4
v = -4e^(-x/4)

Answer 12

integrate by parts again, it becomes:

Answer 13

If again you get
\[-4x^2e^{-x/4}+4[-8xe^{-x/4}+8 \int\limits e^{-x/4}dx]\]

Answer 14

hmmm...

Answer 15

Finally when you've integrated that you'll get
\[\Large [-4x^2e^{-x/4}+4[-8xe^{-x/4}-8(4)e^{-x/4}] ]_0^4\]

Answer 16

do you combine the 1st and the 2nd part, and then apply the limits? or do you only apply limits to the second part?

Answer 17

if you know what i mean?

Answer 18

Yes, use the limits for all

Answer 19

so, i should only apply the limits once ive integrated everything andput the two bits together?

Answer 20

(just to clarify what u mean) :)

Answer 21

Yeah

Answer 22

ahh, ok. yeah, im just integrating now...sorry if im taking a while. im doing it on paper.

Answer 23

how do u know which bit to make u, and which bit to make dv/dx? i know that you have to try and make it simpler

Answer 24

You have to look at it, you know differentiation will reduce powers, and if there's any natural log involved, use u=lnx,

Answer 25

(i love captain jack sparrow :))

Answer 26

yeah ok

Answer 27

nearly there, hang on...

Answer 28

xP

Answer 29

When you have a u that differentiates to zero the tabular method is very handy.

Answer 30

\[-4x ^{2}e ^{-x/4} -8xe ^{-x/4} -32e^{-x/4} \] between 0 and 4

Answer 31

(i will lookup tabular method on google, dw) :p

Answer 32

You forgot to distribute the 4

Answer 33

\[-4x ^{2}e ^{-x/4} -32xe ^{-x/4} -256e^{-x/4}\]

Answer 34

oh yeah, whoops thanks i didnt see that...

Answer 35

After that you should get
\[=64 \left(2-\frac{5}{e}\right)\]

Answer 36

sorry, but how did you get to that exactly please? i am stuck

Answer 37

Get this?
\[-4x ^{2}e ^{-x/4} -32xe ^{-x/4} -256e^{-x/4}\]

Answer 38

yeah. (i forgot to multiply by 4 earlier)

Answer 39

do i take out a factor of 4xe^(-x/4)?

Answer 40

\[\Large [-4x^2e^{-x/4}+\color{blue}{4}[\color{blue}{-8}xe^{-x/4}-\color{blue}{8(4)}e^{-x/4}] ]_0^4\]

Answer 41

not sure what you mean?

Answer 42

sorry. :(

Answer 43

but i will just plug the numbers in anyway...

Answer 44

\[\Large [-4x^2e^{-x/4}+\color{blue}{4}(\color{blue}{-8}xe^{-x/4}-\color{blue}{8(4)}e^{-x/4}) ]_0^4 \\ \\ \Large [-4x^2e^{-x/4}-\color{blue}{32}\color{blue}{}xe^{-x/4}-\color{blue}{256}e^{-x/4} ]_0^4\]

Answer 45

ahhh gotcha

Answer 46

okay, I gotta go, you should get
\[=64 \left(2-\frac{5}{e}\right) \]

Answer 47

yeah, thank you so much for your time and everything :) :) realy appreciate it

Answer 48

ill still be working on it tho.

Answer 49

any new takers? anyone???

Answer 50

hmmm...perplexingggg.....

Answer 51

do i need to take logs?