00:01
Okay, so here we're given a more general function that i need to find the mean and standard deviation.
00:06
So i'm actually given this function right here, 1 over r, e to the negative x over r, dx, right? and i need to find, well, i should also say that the domain is given as that x is from zero to infinity, right? and i need to find the mean, right, which is going to be the integral with one over arm.
00:33
I'm going to take my constant out from zero to infinity in this case of x, e to the negative x over r, dx.
00:43
So for this, i'm going to use integration by parts.
00:45
I'm going to use the tabula method.
00:47
Make my little table here.
00:50
I'm going to integrate my x.
00:52
I'm sorry, i'm going to differentiate my x.
00:54
I'm going to integrate my e to the negative x over r.
00:56
You need x1 ,0, right? here are my signs.
01:04
And then i'm going to integrate e to the negative x over r.
01:07
And i will get, let's see here.
01:12
Negative r, e to the negative x over r.
01:16
And i will get negative r squared.
01:20
Negative r squared.
01:23
E to the negative x over r.
01:25
And i'm going to multiply diagonally.
01:29
So i will get negative r.
01:32
X e to the negative x over r this will be minus r squared e to the negative x over r right and let's not forget my constant out here one over r so what i would do is i would factor an r out here and then combine it with your one over r so you're going to get one over r times i'll factor out actually a negative r and i'll get x e to the negative x over r and that'll be plus r e to the negative x over r, excuse me, and this is from zero to infinity, which of course, this gives us an improper integral, so i need to use a variable and a limit to solve it.
02:16
So i'm going to get the limit.
02:19
I like to use t as t goes to infinity of.
02:25
I'm going to also turn this into a denominator, because it tends to make it easier to see how limits behave as they go to either zero or infinity.
02:34
So i'm going to say t over e to the t over r plus r over e to the t over r minus 0 over e to the 0, which is 1.
02:49
So i'm actually just going to write 1 right here, plus r over e to the 0, which is 1.
02:59
So if i look at this, both top and bottom are going to infinity, so it will be undefined.
03:03
So i need to use lopitals, which means take the derivative of the top and the bottom, derivative of t is one the derivative of course with respect to t of e to the t over r is 1 over r e to the t or r right so as this goes to infinity the entire fraction will go to zero so that goes away this will go to infinity so this entire thing will go to zero this is obviously zero so i'll get negative r but i also have this constant out here which all that is going to reduce to negative 1 so that means that r is my mean.
03:44
So now my standard deviation is given by, i'll take my r out already, 0 to infinity of x minus, whatever my mean is squared, so x minus r squared times my probability density function.
04:04
I already took out my 1 over r, so i'm going to just get e to the negative x over r, dr.
04:12
And now i'm going to, let's see, differentiate this.
04:17
I'm going to use the tabular method again.
04:23
And i'm going to integrate x minus r squared, plus minus and my plus.
04:31
I'll get two, x minus r.
04:33
I'll get two, and i'll get zero.
04:37
And i'm going to integrate e to the negative x over i will get negative r e to the negative x over r you know r squared e to the negative x over r and then i'll get negative r cubed e to the negative x over r and i'm going to multiply diagonally and i will get negative r x minus r squared e to the negative x over r right let's see minus 2r squared x minus r e to the negative x over r and i'll write plus negative r cubed is the same as negative r cubed if that makes sense so i can basically remove the parentheses i'll have negative 2 r cubed e to the negative x over r right um what i would do is i'd factor out a negative r again i'm also going to have my 1 over r out here.
05:51
X minus r squared, e to the negative x over r, plus 2r x minus r, e to the negative x over r.
06:04
And then i will get plus, let's see here...