Question

Integral

Answer 1

?

Answer 2

\[\int\limits_{-\infty}^{\infty}7^{-x^2}dx\]

Answer 3

hint:use normal distribution .. i need help

Answer 4

normal distribution has to deal with statistics

Answer 5

my thought is to find a taylor series for it and bump it up

Answer 6

the normal distribution is the best way to do it

Answer 7

prolly why its a hint :) but i got no clue what that entails

Answer 8

\[f_x(x)=\int\limits_{-\infty}^{\infty}\frac{1}{\sqrt{2 pi}\sigma}e^{\frac{1}{2 \sigma^2}(x- mu)^2}\]

Answer 9

Can someone explain how i can make the question in the form of the normal distribution integral

Answer 10

im sure Zarkon can ;)

Answer 11

doesnt the graph of 7^... look similar to a normal curve?

Answer 12

you forgot a - sign on your normal distribution

Answer 13

yeah \[\large{\int\limits_{-\infty}^{\infty}\frac{1}{\sqrt{2pi} \sigma} e^{-\frac{1}{2 \sigma^2}(x- mu)^2}}dx\]

Answer 14

\[7^{-x^2}=e^{-\ln(7)x^2}=\exp\left(-\frac{x^2}{2\sigma ^2}\right)\]

Answer 15

solve for sigma

Answer 16

then use the fact that \[\int\limits_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi} \sigma} e^{-\frac{1}{2 \sigma^2}(x- mu)^2}dx=1\]

Answer 17

obviously \(\mu=0\)

Answer 18

i dont understand the first part
is it
\[e^{-\ln(7)x^2}=e^{-\frac{x^2}{2 \sigma^2}}\]?

Answer 19

\[\ln(7)=\frac{1}{2\sigma^2}\]

solve for \(\sigma\)

Answer 20

oh ok when i find sigma i just substitute the values in the N.D integral?

Answer 21

sigma can be any positive real number...pick the one that works for your problem

Answer 22

yes

Answer 23

Oh okay,, Thankyou so much Zarkon :D

Answer 24

I get \[\sqrt{\frac{\pi}{\ln(7)}}\]

Answer 25

ok i will try to do it now ,,, i understood the whole idea i just want to try it :)