Question

Ur, another integration-by-parts problem! I have g(a) = the integral from -a to a of e^((-x^2)/9) (x^4)dx. I know two things: I need to integrate twice, and the answer will model the erf(q) equation.

Answer 1

But what I don't get is how to split up the variables to do my intergration-by-parts. Is it u= x^4, dv= e^((-x^2)/2), or vice versa?

Answer 2

either way, I have been struggling with this and not getting a satisfactory answer

Answer 3

Definitely make u equal to the x^4 ! Use the acronym LIATE to figure out what should be u, so that way you don't always guess it out.

L - logarithmic
I - inverse trig
A - algebraic
T - trig
E - exponential

So since algebraic comes before exponential, use the x^4 as your u.

Answer 4

I think we can make use of the same trick we did last time since this function is even

Answer 5

yes we can, we just have to do it twice

Answer 6

Hi guys, just left for a quick sec

Answer 7

So, I was looking at that: can I split the x^4 up? I was trying, and making two intergrations at the moment

Answer 8

but I get\[\int\limits_{-a}^{a}w^2 *e^-w/9\]?

Answer 9

I'm guessing I can do u = w^2, du = 2w; dv = e^(-w/9), v = e^(-w/9) at this point?

Answer 10

that's what I'm trying right now...

Answer 11

\[\Large\int_{-a}^ax^4e^{-\frac{x^2}9}dx=\int_0^ax^3(2xe^{-\frac{x^2}9})dx\]\[u=x^3\implies du=3x^2dx\]\[dv=2xe^{-\frac{x^2}9}dx\implies v=-9e^{-\frac{x^2}9}\]\[\int_0^ax^3(2xe^{-\frac{x^2}9})dx=-9x^3e^{-\frac{x^2}9}+27\int_0^ax^2e^{-\frac{x^2}9}dx\]\[=-9x^3e^{-\frac{x^2}9}+27\int_0^ax(xe^{-\frac{x^2}9})dx\]\[u=x\implies du=dx\]\[dv=xe^{-\frac{x^2}9}dx\implies v=-\frac92e^{-\frac{x^2}9}\]so the integral is now\[-9x^3e^{-\frac{x^2}9}+27(-\frac92xe^{-\frac{x^2}9}+\frac92\int e^{-\frac{x^2}9}dx)\]crap I got stuck...

Answer 12

wait, that's fine I think: I need to integrate twice, and then write it in terms of the erf(q) function I mentioned above. I have similar work like yours, up to a point...

Answer 13

oh you can use the error function ?! I missed that
the last integral is the error function

Answer 14

so then my answer is right :)

Answer 15

Wait, so this is how I reasoned the first substution: did you do something like w = x^2, dw = 2xdx? and then so it would be the integeral 2xwe^-wdx before you did intergration by parts?

Answer 16

no I didn't think like that
I used the even function thing to get it into the form I have on the first line then did u=x^3, dv=2xe^(-x^2/9)

Answer 17

hmm..

Answer 18

I guess I'm having trouble intergrating e^((-x^2)/9) once I try to do the first intergration by parts

Answer 19

\[\Large\int_{-a}^ax^4e^{-\frac{x^2}9}dx=\int_0^ax^3(2xe^{-\frac{x^2}9})dx\]\[u=x^3\implies du=3x^2dx\]\[dv=2xe^{-\frac{x^2}9}dx\implies v=-9e^{-\frac{x^2}9}\]\[\int_0^ax^3(2xe^{-\frac{x^2}9})dx=-9a^3e^{-\frac{a^2}9}+27\int_0^ax^2e^{-\frac{x^2}9}dx\]\[=-9a^3e^{-\frac{a^2}9}+27\int_0^ax(xe^{-\frac{x^2}9})dx\]\[u=x\implies du=dx\]\[dv=xe^{-\frac{x^2}9}dx\implies v=-\frac92e^{-\frac{x^2}9}\]so the integral is now\[-9a^3e^{-\frac{a^2}9}+27(-\frac92ae^{-\frac{a^2}9}+\frac92\int_0^a e^{-\frac{x^2}9}dx)\]\[-9a^3e^{-\frac{a^2}9}+27\left(-\frac92ae^{-\frac{a^2}9}+\frac{27}4\sqrt\pi\text{erf}(\frac a3)\right)\]I'll let you simplify that ;)

Answer 20

okay your problem is that you \(cannot\) integrate\[\int e^{-\frac{x^2}9}dx\]in terms of traditional functions, the answer to the integral is\[\int e^{-\frac{x^2}9}dx=\frac32\sqrt\pi\text{erf}(\frac x3)\]so that is where you are supposed to be stuck :P

Answer 21

right; so I am having issues with doing intergration by parts while being stuck there.

Answer 22

It seems that you don't try to intergrate it at all, you just kind of go with it

Answer 23

you can prove that integral is what it is if you know a bit about the gamma function and laplace transforms

Answer 24

But you still take the intergal of dv...how did you do the integral of 2xe^((-x^2)/9)?

Answer 25

let \[u=e^{-\frac{x^2}9}\implies du=-\frac29xe^{-\frac{x^2}9}dx\]integral becomes\[-9\int du\]

Answer 26

gotta do those simple u-subs in your head as you move through calculus

Answer 27

yea, I get that so far

Answer 28

hold on

Answer 29

\[\int\limits_{}^{} e^(x^2/9) dx\]
= ?
(2x/9)*e^((-x^2)/9)?

Answer 30

I don't think it's as easy as that---or I'm doing something wrong

Answer 31

wait, you don't intergrate/derive the x^2 anymore, only the 1/9 in your work

Answer 32

I am confused what you mean
...btw to keep the exponents where they are supposed to be enclose them in {} not in ()

Answer 33

as I said before you cannot integrate e^(-x^2/9)dx, but you can integrate 2xe^(-x^2/9)dx
the 2x outside allows us to do a simple substitution

Answer 34

oh!

Answer 35

the trick is realizing how similar 2xe^(-x^2/9) is to the derivative of e^(-x^2/9)

Answer 36

I think it clicked, yes?

Answer 37

I'll brb...

Answer 38

I got it! Thanks so much! It finally clicked!

Answer 39

I'm glad, I think I will celebrate this rather difficult integration by parts problem with a beer
see ya!

Answer 40

see yah!

Answer 41

The main trick is to know when to separate the powers of x to fit your purposes, I guess. This will probably appear on an exam.