Integration. (1 Viewer)

[shaon0]

How do you integrate:
x^2*e^-x^2?
I tried to integrate it on a calculator but i got.
0.25*sqrt(pi)*erf(x)-0.5e^-x^2

 

[vds700]

How do you integrate:
x^2*e^-x^2?
I tried to integrate it on a calculator but i got.
0.25*sqrt(pi)*erf(x)-0.5e^-x^2

I may be wrong, but i dont think this is doable, i put it in mathematica and it too comers up with the errf(x) thing. I know for a fact u cant integrate e^x^2, so maybe thats why u cant do this one.

Where did u find this question btw?

 

[tacogym27101990]

umm break it up intox.xe^-x^2?
then use integation by parts?
letting u=x and v'=xe^-x^2
then u'=dx and v=-.5e^-x^2
is you get [(-xe^-x^2)/2] +.5 Se^-x^2 dx

thats all i can think of doing
not sure if it correct tho
this wasnt in a 3 unit text book was it??

EDIT: im obviously wrong if wolfram came up with it being and error

 

[vds700]

umm break it up intox.xe^-x^2?
then use integation by parts?
letting u=x and v'=xe^-x^2
then u'=dx and v=-.5e^-x^2
is you get [(-xe^-x^2)/2] +.5 Se^-x^2 dx

thats all i can think of doing
not sure if it correct tho
this wasnt in a 3 unit text book was it??

EDIT: im obviously wrong if wolfram came up with it being and error

what u did was right, but the problem is u cant integrate e^-x^2

 

[shaon0]

umm break it up intox.xe^-x^2?
then use integation by parts?
letting u=x and v'=xe^-x^2
then u'=dx and v=-.5e^-x^2
is you get [(-xe^-x^2)/2] +.5 Se^-x^2 dx

thats all i can think of doing
not sure if it correct tho
this wasnt in a 3 unit text book was it??

EDIT: im obviously wrong if wolfram came up with it being and error

Yeah it is. Its in the 3unit textbook: 3unit Mathematics Couchman, Jones.
Yeah i did it using i.b.p but i just put it in mathematica to check and i saw that there was an error function.

 

[tacogym27101990]

what was the answer in the back of the book??

 

[shaon0]

you cant integrate e^(-x^2) but there is a special function called the error function erf(x) which you can put in terms of instead.

you can do it by parts, break it up into x and xe^(-x^2), then instead of writing the integral of e^(-x^2) you can put [sqrt(pi).erf(x)/2]

this should then give:


the question should be integrate xe^(-x^2) as this is doable

Ok thanks.
the answer is: 1/3-1/3e.
The Integral above has the limits 0,1.

 

[shaon0]

ah, in that case the question should be integrate x^2 e^(-x^3)

I x^2 e^(-x^3) dx

u = -x^3
du = -3x^2

= (-1/3) I -3x^2 e^(-x^3) dx
= (-1/3) I e^u du (limit change 0-> -1)
= (-1/3)(1/e) + 1/3
= 1/3-1/3e

Bad editing on the book.

 

[tacogym27101990]

so whats the deal with this error function thing 3 unitz??

 

[nottellingu]

This question requires a technique of integration called 'integration by parts', u dont need it for 3u so dont worry about it.

 

[tommykins]

回复: Re: Integration.

This question requires a technique of integration called 'integration by parts', u dont need it for 3u so dont worry about it.

Post the answer please.

 

[doink]

Yep, Taylor series find approximations for functions which you can't integrate then you can integrate them using the taylor series. Think of trying to integrate Arc tan(x), not doable, but with a series in x equating it then its doable.

 

[vds700]

Yep, Taylor series find approximations for functions which you can't integrate then you can integrate them using the taylor series. Think of trying to integrate Arc tan(x), not doable, but with a series in x equating it then its doable.

You can do that using integration by parts can't you?

 

[tacogym27101990]

yeah you can definately integrate atanx using ibp
its just x.atan[x] - .5ln(1+x^2)

 

[vds700]

yeah you can definately integrate atanx using ibp
its just x.atan[x] - .5ln(1+x^2)

yeah, maybe he meant

I(arctanx)^n dx

 

[Js^-1]

Arc tanx means tan^-1x

 

[tommykins]

Yep, Taylor series find approximations for functions which you can't integrate then you can integrate them using the taylor series. Think of trying to integrate Arc tan(x), not doable, but with a series in x equating it then its doable.

x.arctanx - (ln|x²+1|)/2

 

[vds700]

arctan(x) can be written as an infinite series to calculate values (like in your calculator) to a degree of accuracy.

arctan(x) = x - x^3/3 + x^5/5 - x^7/7 + ...

if we wanted a numerical estimate of arctan(0.324235), you could put it into a few terms of the series. you can obviously integrate arctan(x) without any trouble, though if we had limits (such as 0 -> 0.324235) and we wanted a numerical estimate, we would have to use the series.

another interesting thing that results in series representations of functions, is that you can calculate values of known constants such as e and pi to millions of decimal places. think about putting x = 1 into the arctan series before, you will get:

arctan(1) = 1 - 1/3 + 1/5 - 1/7 + ...
pi/4 = 1 - 1/3 + 1/5 - 1/7 + ...
pi = 4 - 4/3 + 4/5 - 4/7 + ...

there are other series which converge even quicker, so you dont need to take as many terms to get a good estimate such as the famous ramanujan pi series. awesome stuff yeah?

thats awesome!

 

[samwell]

this is a 4 unit qsn. integration by parts:
int:x^2*e^-x^2= [uv]-int:vdu given that u=x^2 and dv=e^-x^2
([0.5x*e^-x^2]-int:-e^-x^2) [since du=2x and v=-e^-x^2/2x]
([0.5x*e^-x^2]-[e^-x^2/2x]) [apply limits and then get the answer]

 

[tacogym27101990]

e^-x^2 doesnt integrate to e^-x^2/-2x
didnt you read the other posts?