## Jan 3, 2018

Meanlife $= - \frac{1}{k}$

#### Explanation:

Technically this is a math problem because it involves integration only.

Let

$M = k I$

$I = {\int}_{0}^{\infty} t {e}^{k t} \mathrm{dt}$

Perform integration by parts on I:
$u = t , \mathrm{dv} = {e}^{k t} \mathrm{dt} \Rightarrow v = \frac{1}{k} {e}^{k t}$
$I = u v - v \mathrm{du} = \frac{1}{k} {\left(t {e}^{k t}\right)}_{0}^{\infty} - \frac{1}{k} {\int}_{0}^{\infty} {e}^{k t} \mathrm{dt}$
$I = u v - v \mathrm{du} = \frac{1}{k} {\left(t {e}^{k t}\right)}_{0}^{\infty} - \frac{1}{k} ^ 2 {\left({e}^{k t}\right)}_{0}^{\infty}$

$I = \frac{1}{k} \left[{\left(t {e}^{- | k | t}\right)}_{0}^{\infty} - \frac{1}{k} {\left({e}^{- | k | t}\right)}_{0}^{\infty}\right]$

Because${\lim}_{t \to \infty} t {e}^{- | k | t} \to 0$
For a proof, see https://socratic.org/questions/can-you-find-the-limit-lim-xrarroo-xe-x-2?source=search

$\mathmr{and} k = - | k | \Rightarrow {e}^{- k | t |} \to 0$, when $t \to \infty$

$I = \frac{1}{k} \left(\left(0 - 0\right) - \frac{1}{k} \left(0 - 1\right)\right) = \frac{1}{k} ^ 2$

$M = - k I = - k \left(\frac{1}{k} ^ 2\right) = - \frac{1}{k} = \frac{1}{|} k | = \frac{1}{0.000121}$

M= 8264 yrs

To verify this answer, compute C-14 half-life:

Half-life = $\ln \left(2\right) M = 5729 y r s$

and compare it to the standard C-14 half-life = $5730 \pm 40 y r s$