Continuous Random Variable 
Let X be a random variable with density function fx(.) such that:
  fx(s) = ke$^{-s}$ for s > 2 and fx(s) = 0 for s $\le$ 2. 
1.) Determine k
2.) Calculate P(0 $\le$ X  $\le$ 5).
3.) Calculate E(X)

My attempt:
For 1, I got that $\int_2^\infty k e^{-s}\,ds=2.$ I integrated and got that k = $2e^2$.
For 2, I got that $\int_0^5 k e^{-s}\,ds.$ I integrated and got that $[(2/e^3) -2e^2]$.
For 3, I got that $\int_2^\infty k se^{-s}ds.$ I integrated and got 6.
Is this correct?
 A: $1.$ An antiderivative of $e^{-s}$ is $-e^{-s}$, so 
$$\int_2^\infty ke^{-s}\,ds=ke^{-2}.$$
We want our integral to be $1$, that is always the case for a density. So $k=e^2$.
$2.$ The density function is zero to the left of $2$. So we want
$$\int_2^5 e^2 e^{-s}\,ds.$$
$3.$ We want
$$\int_2^\infty (s)(e^2e^{-s})\,ds.$$
Now do integration by parts. Note again that the density function is $0$ to the left of $2$, hence the limits.  Your answer, which is negative (impossible) seems to have been obtained by integrating from $0$.  
Now do integration by parts. It looks as if you got this right, apart from having a $k$, and hence an answer, twice as lage as it should be.
Remark: There is an (in this case) sloghtly easier way to calculate the expectation, that you may not have been taught. If $X$ is non-negative, with *cumulative distribution function $F_X(s)$, then 
$$E(X)=\int_0^\infty(1-F_X(s))\,ds.$$
In our case, we would really be integrating from $2$ to infinity, since the cdf is $0$ to the left of $2$.
A: By definition $$ f_X(x) = \cases{ k \exp(-x) & $x>2$ \\ 0 & $x \leqslant 0$}$$
The normalization constant is determined by requiring $1 = \int_\mathbb{R}f_X(x) \mathrm{d}x$:
$$
  1 = \int_\mathbb{R}f_X(x) \mathrm{d}x = \int_2^\infty k \exp(-x) \mathrm{d}x \stackrel{x=2+y}{=} k \mathrm{e}^{-2} \underbrace{ \int_0^\infty \exp(-y) \mathrm{d}y}_{ = 1}
$$
hence $k = \mathrm{e}^2$. Thus:
$$
    f_X(x) = \cases{ \exp(-(x-2)) & $x>2$ \\ 0 & $x \leqslant 0$}
$$
that is $X= 2+Y$, where $Y$ is the exponential random variable with unit mean. Thus
$$
  \mathbb{P}\left(0\leqslant X\leqslant5\right) = \mathbb{P}\left(-2\leqslant Y\leqslant3\right) = \mathbb{P}\left(0\leqslant Y\leqslant3\right) = 1 - \underbrace{\mathbb{P}(Y>3)}_{\int_3^\infty \exp(-t) \mathrm{d}t = \exp(-3)} = 1 - \exp(-3)
$$
The mean is also easy:
$$
 \mathbb{E}(X) = \mathbb{E}(2+Y) = 2 + \mathbb{E}(Y) = 2 + 1 =3
$$
