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anonymous 3 years ago Calculus: Find K

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1. anonymous

|dw:1353910548451:dw|

2. anonymous

You need to use integration by parts

3. hartnn

put u= -y^2/2 then.

4. anonymous

Is it:$\int\limits\limits_{-\infty}^{\infty}ke^{-\frac{y^2}{2}}dy=1$?

5. anonymous

Yes

6. hartnn

u= -y^2/2 find dy =... ?

7. anonymous

I used u=y^2

8. anonymous

but maybe its easier with u=-y^2/2

9. anonymous

Its not smart to have that negative there

10. anonymous

Do you agree?

11. anonymous

@hartnn ?

12. hartnn

oh, sorry.... yeah, its easier with e^u constants can be rearranged easily...

13. anonymous

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.

14. anonymous

ohhhhh

15. anonymous

Yes the answer is 1/ sqrt(2*pi) I just so dont follow how they got that answer.

16. hartnn

did u try that substitution and then integration by parts ?

17. anonymous

I was getting weird answers. I have not really touched calculus for over a yr so i cld be messing up

18. hartnn

$$\huge\int \frac{e^{-u}u^{-1/2}}{\sqrt2}du$$ when 2u=y^2 yes, its messy...

19. anonymous

How would you continue from here?

20. anonymous

Can you show me?

21. anonymous

ok dont show me

22. hartnn

u need to know gamma function to solve that ... i am so sorry, my internet are giving me connection problems....

23. anonymous

hmmm I know the gamma distribution function

24. anonymous

ohhhh I see it now lol

25. hartnn

$$\Gamma(z) = \int_0^\infty e^{-t} t^{z-1} dt$$

26. anonymous

Yesssssss

27. hartnn

now , you'll get it easily.

28. hartnn

ask if u don't get...

29. anonymous

Thanks let me try to work it out then. Thanks :)))))

30. anonymous

what?

31. hartnn

$$\Gamma{1/2} = \sqrt{\pi}$$

32. hartnn

you'll need it.

33. anonymous

ohhh coool never knew that

34. anonymous

I feel like i am missing the basics in math :(

35. hartnn

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