Processing...

33.5 Applications of calculus by Benjamin Crowell, Light and Matter licensed under the Creative Commons Attribution-ShareAlike license.

The area under the probability distribution is of course an integral. If we call the random number `x` and the probability distribution `D(x)`, then the probability that `x` lies in a certain range is given by

`("probability of a"<="x"<="b")=int_a^bD(x)dx`.

What about averages? If `x` had a finite number of equally probable values, we would simply add them up and divide by how many we had. If they weren't equally likely, we'd make the weighted average `x_1P_1+x_2P_2+...` But we need to generalize this to a variable `x` that can take on any of a continuum of values. The continuous version of a sum is an integral, so the average is

`("average value of x")=intxD(x)dx,`

where the integral is over all possible values of `x`.

Here is a rigorous justification for the statement in section 33.4 that the probability distribution for radioactive decay is found by substituting `N(0)=1` into the equation for the rate of decay. We know that the probability distribution must be of the form

`D(t)=k0.5^(t"/"t_(1"/"2))`,

where `k` is a constant that we need to determine. The atom is guaranteed to decay eventually, so normalization gives us

`("probability of 0"<="t"<infty)=1`

`=int_0^inftyD(t)dt.`

The integral is most easily evaluated by converting the function into an exponential with `e` as the base

`D(t)=kexp[ln(0.5^(t"/"t_(1"/"2)))]`

`=kexp[t/t_(1"/"2)ln0.5]`

`=kexp(-(ln2)/t_(1"/"2)t),`

which gives an integral of the familiar form `inte^(cx)dx=(1/c)e^(cx)`. We thus have

`1=-(kt_(1"/"2))/(ln2)exp(-(ln2)/t_(1"/"2)t),`

which gives the desired result:

`k=(ln2)/t^(1"/"2)`

You might think that the half-life would also be the average lifetime of an atom, since half the atoms' lives are shorter and half longer. But the half whose lives are longer include some that survive for many half-lives, and these rare long-lived atoms skew the average. We can calculate the average lifetime as follows:

`("average lifetime")=int_0^inftytD(t)dt.`

Using the convenient base `-e` form again, we have

`("average lifetime")=(ln2)/t_(1"/"2)int_0^inftytexp(-(ln2)/t_(1"/"2)t)dt,`

This integral is of a form that can either be attacked with integration by parts or by looking it up in a table. The result is `intxe^(cx)dx=x/ce^(cx)-1/c^2e^(cx)`, and the first term can be ignored for our purposes because it equals zero at both limits of integration. We end up with

`("average lifetime")=ln2/t_(1"/"2)(t_(1"/"2)/ln2)^2`

`=t_(1"/"2)/ln2`

`=1.443t_(1"/"2)`,

which is, as expected, longer than one half-life.

33.5 Applications of calculus by Benjamin Crowell, Light and Matter licensed under the Creative Commons Attribution-ShareAlike license.