integral question...?

Question

Given integral from -infinity to +infinity of e^(-x^2)dx = sqrt pi,
1. calculate exact value of integral from -infinity to +infinity [e ^ -(x-a)^2 / b ] dx
2. the kth moment, m(sub k) of the normal distribution is defined by integral from -infinity to +infinity [ (x^k) (e ^ (-x^2 / 2) ] dx. Find m (sub2) and m (sub4).

Looks sorta confusing..if you have questions about how i wrote the problem, just leave a msg.

Answer 1

Hi,

This not just ...sqrt(pi) as one respondent thought.

What you need are two very useful tools for integration, part one is solved by changing variables and part two is solved with integration by parts.

1. let x= (y-a)/sqrt(b), so dy/dx = sqrt(b)
=> dx = (1/sqrt(b)) dy (this isn't strictly correct but it works here)

so:
I = int(exp(-(x-a)^2/b)dx,-inf..+inf)
I = int(exp(-((x-a)/sqrt(b))^2)dx,-inf..+inf)
I = int(exp(-y^2)*(1/sqrt(b))dy,-inf..+inf)
I=(1/sqrt(b))*int(exp(-y^2)dy,-inf..+inf)
I=(1/sqrt(b))*sqrt(pi) (from our given result)
I = sqrt(pi/b)

The seond part requries a decent understanding of integration by parts, and if you email me I can go through this with you.

Answer 2

1) the same. still sqrt pi, because the graphp is shifted and still integhrated over an infinite inteval.
2) todelidokie i am not sloving that for you, if your scholl is asking these questions then you should be capable of resolving this

Answer 3

Much more interesting than HOW you wrote the problem is WHY.

Answer 4

Hint: use polar coordinates