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Calculus: Find K

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You need to use integration by parts
put u= -y^2/2 then.

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Other answers:

Is it:\[\int\limits\limits_{-\infty}^{\infty}ke^{-\frac{y^2}{2}}dy=1\]?
u= -y^2/2 find dy =... ?
I used u=y^2
but maybe its easier with u=-y^2/2
Its not smart to have that negative there
Do you agree?
oh, sorry.... yeah, its easier with e^u constants can be rearranged easily...
I got the wrong answer when evaluating the integral, and I won't do it again now. But this function is a gaussian, and it is expected that the constant k in this case has the value 1/sqrt(2pi)=sqrt(2pi)/2pi. That comes from the statistic theory.
Yes the answer is 1/ sqrt(2*pi) I just so dont follow how they got that answer.
did u try that substitution and then integration by parts ?
I was getting weird answers. I have not really touched calculus for over a yr so i cld be messing up
\(\huge\int \frac{e^{-u}u^{-1/2}}{\sqrt2}du\) when 2u=y^2 yes, its messy...
How would you continue from here?
Can you show me?
ok dont show me
u need to know gamma function to solve that ... i am so sorry, my internet are giving me connection problems....
hmmm I know the gamma distribution function
ohhhh I see it now lol
\( \Gamma(z) = \int_0^\infty e^{-t} t^{z-1} dt \)
now , you'll get it easily.
ask if u don't get...
Thanks let me try to work it out then. Thanks :)))))
\(\Gamma{1/2} = \sqrt{\pi}\)
you'll need it.
ohhh coool never knew that
I feel like i am missing the basics in math :(

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