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Integration by Parts 5

https://isaacphysics.org/questions/integration_parts5?board=c1379f09-5ce8-460d-8ad3-0c5fd7492a30&stage=a_level

Please could someone tell me if my working steps so far are correct and if so, what does the term on the left approach as x->+/- infinity
Working:
parts.jpg
Reply 2
Original post by mosaurlodon
Working:
parts.jpg

Why not think a bit more (try a line or two) for which terms to integrate and which terms to differentiate? If you integrate the x^2 term, the exponent is only going to keep get larger when you do by parts again.

A hint is that x^2 = x*x.

The term on the left would approach 0, why not try plotting it in desmos or ... and thinking about why.
(edited 1 month ago)
@mosaurlodon Just to say that although I think mqb is right to just give a hint here, the correct way to proceed is quite hard to spot if you've never seen it before, so ask for more help if you need it.
Ok I get why the left part is 0, if you sub infinity for x in e^-x^2... =0 and so that cancels with x^3, and vice versa if you sub in negative infinity.

I think I kind of understand the trick here?
Im assuming we want to get rid of the x's so you choose u=x, so du/dx = 1, and I just integrated the other part, and somehow the x also got removed - so I think im on the right track?
integral.png
Reply 5
Original post by mosaurlodon
Ok I get why the left part is 0, if you sub infinity for x in e^-x^2... =0 and so that cancels with x^3, and vice versa if you sub in negative infinity.
I think I kind of understand the trick here?
Im assuming we want to get rid of the x's so you choose u=x, so du/dx = 1, and I just integrated the other part, and somehow the x also got removed - so I think im on the right track?
integral.png

You probably want to be a bit clearer about why it "cancels" the x^3. Simply put is that an exponential "beats" a polynomial so it grows or decays faster. Here you an exponential with a square so decays faster than an e^(-x). A bit more formally you could write
x^3 e^(-x^2)
as using the exponential power / maclaurin series
x^3 / (1 + x^2 + x^4/2 + x^6/6 + ...) = 1/(1/x^3 + 1/x + x/2 + x^3/6 + ....)
so as x gets large (positive or negative) the denominator -> +/-inf and the fraction ->0. This obviously applies to the corrected working as youre considering x e^(-x^2).

For the by parts, youre doing it the right way so roughly you have
x^2 e^(-x^2)
We want the x^2 term to decrease so differentiate it. But that requires integrating e^(-x^2) which isnt an elementary result. But we notice that differentiating it we get roughly x e^(-x^2). So we can choose to differentiate x and integrate xe^(-x^2) and things work out.

Sometimes its easier to sketch the working without worrying about extra constants etc to make sure the basic logic is sound, then do the actual working with all the constants etc. It looks about right with your working, just write the final line.
(edited 1 month ago)
Ohhh that logic makes so much sense... I just got lucky with the way I choose to integrate.

I was actually doing integrating by parts again until you said 'final line' then I realised...

thanks for your help :smile:
Reply 7
Original post by mosaurlodon
Ohhh that logic makes so much sense... I just got lucky with the way I choose to integrate.
I was actually doing integrating by parts again until you said 'final line' then I realised...
thanks for your help :smile:

No problem. For the final line part, thats why Id suggest doing a simpler by parts first, without all the constants etc. If youre aiming to differentiate the x^2 (roughly) so that youre just left with a definite integral of e^(-x^2), then you can do that because youre given it in the question. Otherwise its a hard thing to integrate. That really is the aim of the by parts and if youre not clear about that at the start, then as you say, it can amount to luck.

A slight aside, but youre really calculating E(x^2) for a gaussian distribution and its simply the variance so sigma^2.

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