/sci/ - Science & Math » Thread #11195955

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Anonymous Tue 03 Dec 16:27:41 2019 No.11195955 View Reply Original Report

Quoted By: >>11195966 >>11195981 >>11196158 >>11196162 >>11196373 >>11196466

Can someone please help me with this?

Anonymous

Anonymous Tue 03 Dec 2019 16:33:32 No.11195966 Report

Quoted By: >>11195972

>>11195955
You can't even make a provocative post to disguise that you want help with homework.

Anonymous

Anonymous Tue 03 Dec 2019 16:36:10 No.11195972 Report

Quoted By: >>11196024 >>11196158

>>11195966

Not really I work in like 2 hours and want to get this problem solved so I can ponder over it while working instead of being pissed about it while I work because it’s unfinished :(

Anonymous

Anonymous Tue 03 Dec 2019 16:40:17 No.11195981 Report

Quoted By: >>11196158

>>11195955
Just put it into wolfram or symbolab instead of asking here retard

Anonymous

Anonymous Tue 03 Dec 2019 17:07:20 No.11196024 Report

Quoted By: >>11196158

>>11195972
Isn’t this a basic application of gamma functions

Anonymous

Anonymous Tue 03 Dec 2019 17:11:57 No.11196031 Report

Quoted By: >>11196158

literally just the fundamental theorem of calculus

Anonymous

Anonymous Tue 03 Dec 2019 18:02:22 No.11196120 Report

Quoted By: >>11196158

it's literally just an aplication of the fundamental theorem of calculus.
You just need to invert the limits on the first integral and it's done....

Phil F

Phil F Tue 03 Dec 2019 18:25:12 No.11196158 Report

Quoted By:

>>11195955

>>11195981
>>11195972
>>11196024
>>11196031
>>11196120
None of these guys have any idea...
Paypal me or find me on Chegg- Phil F.

Anonymous

Anonymous Tue 03 Dec 2019 18:27:11 No.11196162 Report

Quoted By:

>>11195955
It’s literally just zero

Anonymous

Anonymous Tue 03 Dec 2019 20:04:15 No.11196373 Report

Quoted By:

>>11195955
2e^(-x^2)

Anonymous

Anonymous Tue 03 Dec 2019 20:58:31 No.11196466 Report

Quoted By:

>>11195955
Assume there exists some g(s) such that g'(s) = e^(s^2). The integral from -x to x of g'(s) is just g(x) - g(-x) = 2g(x). If we then take the derivative of this, we get d(2g(x))/dx = 2g'(x) = 2e(x^2). I think that works

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