Finding the Expected Value and Variance The distribution function of a random variable $X$ is given by
$P (X \leq x)$ =
\begin{cases}
0 & \text{if $x<0$}, \\
x^{k} & \text{if $0 \leq x \leq 1, \,\,\,\,\,\, k\geq1$}, \\
1     & \text{if $x > 1$}.
\end{cases}
Determine the mean and variance of $X$.
Workings:
I am aware that I must start by calculating $\sum_{-\infty}^{\infty} x \,\cdot p_X(x)$, but because this is a continuous distribution, I will instead have to calculate $\int_{-\infty}^{\infty} x \, \cdot p_X(x) \,dx$. However, once I integrate it, I am confused as to what further steps as to which I am meant to take given that I get infinity as the expected value. Could someone please tell me if it is possible for the expected value to be infinite and if I have made a mistake in my workings? Thank you.
\begin{align}
\text{E(X)} & = \int_{-\infty}^{\infty} x \, \cdot p_X(x) \,dx \\
            & = \int_{-\infty}^{0} x \, \cdot 0 \,dx \: + \int_{0}^{1} x \, \cdot x^{k} \,dx \: + \int_{1}^{\infty} x \,dx \\
& = C + \bigg{[}\frac{x^{k+2}}{k+2}\bigg{]}_{0}^{1} + \bigg{[}\frac{x^{2}}{2}\bigg{]}_{1}^{\infty}
\end{align}
 A: You are using the CDF in your integral, and you should instead use the PDF, which is the derivative $$p_X(x) = \begin{cases}0 & x \le 0 \\ k x^{k-1} & 0 \le x \le 1 \\ 0 & x > 1\end{cases}$$
A: No reason to compute the PDF here. If $X$ is a nonnegative random variable (w.p.1) which it is in this case, you can calculate:
$$\mathbb{E}[X] = \int_0^\infty \mathbb{P}(X \geq x)\mathrm{d}x =\int_0^\infty (1- F_X(x)\mathrm{d}x = \int_0^1 1-x^k \mathrm{d}x + \int_1^\infty 1-1\mathrm{dx} = \int_0^1 (1-x^k)\mathrm{d}x$$
For the variance, we just need to compute:
$$
\mathbb{E}[X^2] = \int_0^\infty \mathbb{P}(X^2 \geq x)\mathrm{d}x = \int_0^\infty \mathbb{P}(X \geq \sqrt{x} \text{ or } X \leq -\sqrt{x} )
$$
Because $X$ is nonnegative, we have:
$$
\mathbb{E}[X^2] = \int_0^\infty \mathbb{P}(X \geq \sqrt{x})\mathrm{d} x = \int_0^\infty
(1- F_X(\sqrt x))\mathrm{d}x = \int_0^1 1- x^{k/2} \mathrm{d}x$$
Once you have this the variance is:
$$
\mathbb{E}[X^2] - \mathbb{E}[X]^2
$$

To prove the formula, let $X$ be a nonnegative random variable with density/PDF $f_X$. Note that:
$$\mathbb{P}(X \geq x) = \int_x^\infty f_X(y) \mathrm{d}y$$
then:
$$
\int_0^\infty \mathbb{P}(X \geq x) \mathrm{d}x = \int_0^\infty \int_x^\infty f_X(y) \mathrm{d}y\mathrm{d}x
$$
This is a double integral over the set in the plane given by $\{(x,y) \in \mathbb{R}^2 : x > 0 , y>x\}$ (i.e. an infinite triangle) of the function $g(x,y) = f_X(y)$ (only a function of $y$).
We can use Fubini's theorem (or more technically, Tonelli's theorem if you wish) to switch the order of integration and integrate with respect to $x$ first and $y$ second. This then becomes:
$$
\int_0^\infty \int_x^\infty f_X(y) \mathrm{d}y\mathrm{d}x = \int_0^\infty\int_0^y f_X(y)\mathrm{d}x\mathrm{d}y = \int_0^\infty yf_X(y) \mathrm{d}y
$$
which is the formula for $\mathbb{E}[X]$.
This formula holds in more generality, for example for discrete RVs and continuous RVs that don't have densities.
A: You are integrating the distribution function and not the probability density
https://en.wikipedia.org/wiki/Probability_density_function
The two are related because the distribution function gives you the probability that $X$ is in $]-\infty,x]$ and the integral of the probability density between $a$ and $b$ gives you the probability of finding $X$ in $[a,b]$, so the distribution function $P(X\le x)$ is related to the probability density $p_X(x)$ by the relation
$$P(X\le x)=\int_{-\infty}^xp_X(t)dt$$
A: 
$$ \int_{-\infty}^{0} x \, \cdot 0 \,dx+\ldots \\ =C+\ldots$$

A side note to this part: If you have a definite integral you don´t have the constant C as a part of the result.
$$\int_{a}^b f(x) \ dx=F(b)-F(a)$$
That means in the interval $(-\infty, 0)$ your integral is
$$\int_{-\infty}^0 x\cdot 0 \ dx=0\cdot \int_{-\infty}^0 x \ dx=0\cdot \left[ \frac12\cdot x^2 \right]_{-\infty}^0=0-0=0$$
The last equality holds since one factor is zero:
$"\textrm{If at least one factor of a product is zero, then the entire product is zero.}"$
A: Although this can be solved by direct integration, it would be easier if we notice that $X$ follows a $\text{Beta}(k,1)$ distribution:
$$f_X(x)=kx^{k-1}\propto x^{k-1}$$
We can directly see that $$\mathbb E(X)=\frac{k}{k+1},\:\text{Var}(X)=\frac{k}{(k+1)^2(k+2)}$$
By the way it isn't uncommon for the expected value to be undefined (see Cauchy distribution), but in this case it is well defined.
