Determine F(3) and F(1/$\lambda$) from a frequency function I have this problem:
The life span $\xi$ in a radioactive atom has the frequency function:
$$f(x)=\begin{cases}\lambda e^{-\lambda x}&x>0\\
0&x < 0\end{cases}$$ 
Determine $F(3)$ and $F(1/\lambda)$
I don't know how to do this. It seems like it creates an integral going from 0 to infinity but I got stuck there.
The answer is F(3) = $1-e^{-3\lambda}$ and F($1/\lambda$) = 0.63
 A: Note that in this case 
$$
F(x)=P(X\leq x)=\int_{0}^x f(t) \,dt.
$$
In particular
$$
F(3)=\int_{0}^3\lambda e^{-\lambda x}\, dx.
$$
Hopefully you can compute this integral.
A: I assume in the above that you are using the term frequency function to describe a probability density function, and that $F$ is the associated cumulative distribution function.
In this case
\begin{align*}
F(x) & =\mathbf{P}[ \xi \leq x] \\
& = \int_{0}^x \lambda e^{-\lambda y} dy \\
& = \lambda \left[ -\frac{1}{\lambda}e^{-\lambda y} \right]_{0}^x \\
& = 1 - e^{-\lambda x}.
\end{align*}
Hence we have
$$F(3) = 1 - e^{-3 \lambda}$$
whilst
$$F(1/\lambda) = 1 - e^{-1} \approx 0.63$$

Computing the anti-derivative of $e^{\alpha x}$
In the above, I assumed the indefinite integral formula 
$$ \int e^{\alpha y} d y = \frac1\alpha e^{\alpha y},$$
for general $\alpha \in \mathbf{R}$. Probably the easiest way to justify this is by arguing that integration is the `opposite' of differentiation. 
If you are happy with the formula that
$$ \frac{d}{dx} e^{\alpha x} = \alpha e^{\alpha x}$$
then (very informally)
$$ \int \alpha e^{\alpha x} = \int \frac{d}{dx} e^{\alpha x} = C + e^{\alpha x},$$
justifying the formula for the indefinite integral. To make this argument formal you would need to use the Fundamental Theorem of Calculus.
