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Hars integration

Hard*

I can not do.

Find c in

c ∫ exp ( -(2x^2+3x+1)) dx =1, from -infinity to +infinity
Reply 1
Avatar for TeeEm
TeeEm
19
Original post by AlmostNotable
Hard*

I can not do.

Find c in

c ∫ exp ( -(2x^2+3x+1)) dx =1, from -infinity to +infinity


this undergrad standard
substitution throws this I think into exp (-x2)
I can look at this later tonight or tomorrow morning
Original post by TeeEm
this undergrad standard
substitution throws this I think into exp (-x2)
I can look at this later tonight or tomorrow morning


Okay I will go back to figuring this can opener out. Real puzzle.
Reply 3
Avatar for TeeEm
TeeEm
19
Original post by AlmostNotable
Okay I will go back to figuring this can opener out. Real puzzle.


It is done the way I suggested ...
Not too bad, assuming you are an undergrad.
Original post by TeeEm
It is done the way I suggested ...
Not too bad, assuming you are an undergrad.


I am currently doing Alevels.
Reply 5
Avatar for Kummer
Kummer
7
Completing the square we have:

∫−∞∞e−(2x2+3x+1)  dx=∫−∞∞e−2(x+34)2+18  dx=e1/8∫−∞∞e−2(x+34)2  dx\displaystyle \int_{-\infty}^{\infty} e^{-(2x^2+3x+1)}\;{dx}= \int_{-\infty}^{\infty} e^{-2(x+\frac{3}{4})^2+\frac{1}{8}} \;{dx} = e^{1/8}\int_{-\infty}^{\infty} e^{-2(x+\frac{3}{4})^2}\;{dx}

Let t=2(x+34)t = \sqrt{2}\left(x+\frac{3}{4} \right), then it's e1/82∫−∞∞e−t2  dt=e1/8π2\displaystyle \frac{e^{1/8}}{\sqrt{2}}\int_{-\infty}^{\infty}e^{-t^2}\;{dt} =\frac{e^{1/8}\sqrt{\pi}}{\sqrt{2}}
(edited 8 years ago)
Reply 6
Avatar for TeeEm
TeeEm
19
Original post by AlmostNotable
I am currently doing Alevels.


where does the integral come from?
Original post by TeeEm
where does the integral come from?


The cyberspace.
Reply 8
Avatar for TeeEm
TeeEm
19
Original post by AlmostNotable
The cyberspace.


I do not follow but this is undergrad stuff.
here is my version, but this is beyond A level
IMG_0001.jpg387.5KB

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