Answer
M = $\dfrac{-1}{k}$ with $k$ is radioactivity constant
Work Step by Step
We have: M = $-k\displaystyle\int\limits_{0}^{\infty} t.e^{kt}dt = -\dfrac{1}{k} \displaystyle\int\limits_{0}^{\infty} kt.e^{kt}d(kt) $.
After that, we call $ x = kt $. Because $t$ (time) $\gt0$ and $k \lt0$, $x$ must negative. When $t=\infty, x=-\infty$, when $t=0, x=0$.
Using the Definition of Improper Integral of Type 1,
$M= -\dfrac{1}{k} \displaystyle\int\limits_{0}^{-\infty} x.e^{x}dx
= -\dfrac{1}{k}. \displaystyle \lim_{a \to -\infty}\int\limits_{0}^{a} x.e^{x}dx
= -\dfrac{1}{k}.\displaystyle\lim_{a \to -\infty}\left( x.e^{x} |_{0}^{a} -\int\limits_{0}^{a} e^{x}dx\right)$
$
= -\dfrac{1}{k}.\displaystyle\lim_{a \to -\infty}\left[a.e^{a} -(e^{a}-1)\right]
= -\dfrac{1}{k}.\displaystyle\lim_{a \to -\infty}\left[(a-1)e^{a}+1\right]. $
We know $\displaystyle\lim_{a \to -\infty}(a-1)e^{a}=0$ because $\lim\limits_{a \to -\infty}e^{a}=e^{-\infty}=\dfrac{1}{e^{\infty}}=0$.
Therefore M$= -\dfrac{1}{k}.(0+1)=-\dfrac{1}{k} \approx 8264.46 $.
The radiation time of $^{14}$C radioisotope is year so M$ =8264.46$ years.
