Question

Check my work please .. Normal distribution integral

Answer 1

find \[\large{\int\limits_{-\infty}^{\infty}e^{-\frac{1}{2}x^2-3x+6}dx}\] this is what i did\[\large{\int\limits_{-\infty}^{\infty}e^{-\frac{1}{2}(x-3)^2+\frac{3}{2}}dx}\]\[e^{\frac{3}{2}}\int\limits_{-\infty}^{\infty}e^{-\frac{1}{2}(x-3)^2}dx\]
X:N(3,1)
\[\large{=e^{\frac{3}{2}}\int\limits_{-\infty}^{\infty}\frac{\sqrt{2 pi}}{\sqrt{2 pi}}e^{-\frac{1}{2}(x-3)^2}dx}\]\[\large{=\sqrt{2 pi}e^{\frac{3}{2}}\int\limits_{-\infty}^{\infty}\frac{1}{\sqrt{2pi}}e^{-\frac{1}{2}(x-3)^2}dx}\]so integral \[=\large{e^{\frac{3}{2}} \sqrt{2pi}}\]

Answer 2

And please if i have something like\[\large{\int\limits_{-\infty}^{\infty}x^2e^{-x^2}dx}\]how do i start? just give me like a hint

Answer 3

\[-\frac{1}{2}x^2-3x+6=-\frac{(x+3)^2}{2}+\frac{21}{2}\]

Answer 4

integration by parts

Answer 5

oh i messed up with negative ... for the second one can it be solved using N.d or not? just so i can know
Thanks again

Answer 6

yes...use integration by parts and the ND

Answer 7

ok perfect,,, U are the best:)

Answer 8

\[\int\limits_{-\infty}^{\infty}x^2e^{-x^2}dx\]
i am not able to do it :S
its not working byparts

Answer 9

\[\int\limits_{-\infty}^{\infty}x^2e^{-x^2}dx=\int\limits_{-\infty}^{\infty}(x)\left(xe^{-x^2}\right)dx\]

let \(u=x\) and \(dv=xe^{-x^2}dx\)

Answer 10

@Zarkon can i do it like this
\[\int\limits\limits_{-\infty}^{\infty}\frac{\sqrt{2 pi}}{\sqrt{2pi}}\times \frac{\sqrt{\frac{1}{2}}}{\sqrt{\frac{1}{2}}}x^2 e^{-x^2}dx\]\[=\sqrt{2pi}\times \sqrt{\frac{1}{2}}\int\limits_{-\infty}^{\infty}\frac{x^2}{\sqrt{2pi}\times \sqrt{\frac{1}{2}}}e^{-x^2}dx\]the integral now is E[x^2]
and we know that var(x)=E[x^2]-(E[x])^2
so here sigma=sqrt(1/2)
and mu=0
X:N(0,sqrt(1/2))
so var(x)=sigma^2
1/2=E[x^2]-0
that means that \[\int\limits_{-\infty}^{\infty}x^2e^{-x^2}dx=\frac{1}{2} \times \sqrt{2pi}\times \frac{1}{\sqrt{2}}\]
is this right?

Answer 11

yes

Answer 12

though you should simplify it and use \pi in your \(\LaTeX\) for \(\pi\)

Answer 13

yes that is right answer, little unfamilar with how you did it

integration by parts will work as well
you can integrate \[\int\limits_{}^{}xe^{-x^{2}} dx\] using substitution
u = -x^2
du = -2x

and then the definite integral \[\int\limits_{-\infty}^{\infty}e^{-x^{2}} dx = \sqrt{\pi }\]

Answer 14

yeah i did that i was just learning other methods too :D so if i get stuck one id use the other... Thanks alot dumbcow :D

@zarkon ill keep that in mind,,, i dont use latex verymuch so thankyou for the tip:D