I forgot to mention that I am trying to find sigma squared.. given that the mean is r and p(x)= (1/r)(e^(-x/r)) on the interval [0,infinity), where r>0..
note: sigma squared= integral sign (-inf, inf) (x-M)^2 p(x) dx
i got (r^2 - 2r)^(1/2) for sigma, but my answer is wrong...
Previous Posts:
I tried to integrate this:
intergral sign with an interval from -inifinity to + infinity
function: (x-r)^2 (1/r) (e^(-x/r))
with respect to dx.....
I know that Steve set up the u-substitution but I seem to never arrive at an answer... It seems like I keep integrating and integrating and integrating and never arrive at an answer!!!!!!!!
Steve gave you:
let
u = (x-r)^2
du = 2(x-r) dx
dv = (1/r) e^(-x/r) dx
v = e^(-x/r)
CHECK SIGN
∫ u dv = uv - ∫ v du
= (x-r)^2 e^(-x/r) - 2∫(x-r) e^(-x/r) dx
now the second part of it. He said to do it again.
u = x - r
du = dx
dv = e^-x/r
v = -r e^-x/r
-(x-r)e^-x/r - ∫-re^-x/r dx
now you can do ∫-re^-x/r dx = (1/r)e^-x/r
CHECK THE SIGNS !
then put it back together and do the limits
To find sigma squared given the probability density function p(x) = (1/r)(e^(-x/r)) on the interval [0, infinity), you can use the formula:
sigma squared = ∫(x - r)^2 * p(x) dx
Using the previous work with u-substitution, we have:
∫(x - r)^2 * (1/r) * e^(-x/r) dx
First, let's review the steps you've done so far:
1. Set u = (x - r)^2 and dv = (1/r) * e^(-x/r) dx.
2. Calculate du = 2(x - r) dx and v = ∫(1/r) * e^(-x/r) dx = -e^(-x/r).
3. Apply the integration by parts formula: ∫u dv = uv - ∫v du.
Plugging in the values we found:
∫(x - r)^2 * (1/r) * e^(-x/r) dx = (x - r)^2 * (-e^(-x/r)) - ∫(-e^(-x/r)) * 2(x - r) dx
Simplifying this expression:
= -(x - r)^2 * e^(-x/r) + 2 ∫(x - r) * e^(-x/r) dx
Now we can proceed to the second part of the integration by parts:
1. Set u = x - r and dv = e^(-x/r) dx.
2. Calculate du = dx and v = -r * e^(-x/r).
3. Apply the integration by parts formula: ∫u dv = uv - ∫v du.
Plugging in the values we found:
-(x - r) * e^(-x/r) - ∫(-r * e^(-x/r)) * dx
Simplifying this expression:
= -(x - r) * e^(-x/r) + r ∫e^(-x/r) dx
The integral of e^(-x/r) dx can be solved straightforwardly:
= -(x - r) * e^(-x/r) + r * (-r * e^(-x/r)) + C
Now let's combine the two parts and evaluate the expression at the limits [0, infinity):
= -(x - r)^2 * e^(-x/r) + 2 * (-(x - r) * e^(-x/r) + r * (-r * e^(-x/r))) | [0, infinity)
Taking the limit as x approaches infinity:
= -∞ - r^2 * e^(-∞) + 2 * (-(∞ - r) * e^(-∞) + r * (-r * e^(-∞)))
Evaluating the limit as x approaches 0:
= -(0 - r)^2 * e^(0/r) + 2 * (-(0 - r) * e^(0/r) + r * (-r * e^(0/r)))
Simplifying these expressions:
= -r^2 + 2(r - r^2) | [0, infinity)
= -r^2 + 2r - 2r^2 - (-r^2 + 2r - 2r^2) | [0, infinity)
= -r^2 + 2r - 2r^2 + r^2 - 2r + 2r^2
= -r^2 + r^2
= 0
Therefore, sigma squared is equal to zero.