Did I do this Continuous Probability Problem Correctly? I'm new to evaluating continuous probability density functions. I'd like someone to check my work, please.
Problem:

Suppose $X$ has density $f(x) = c/x^6$ for $x>1$ and $f(x) = 0$
  otherwise, where $c$ is a constant.
  
  
*
  
*Find $c$.
  
*Find $E(X)$.
  
*Find $Var(X)$.
  

Solution:


*

*The big rule we'll refer to is $\int_{-\infty}^\infty f(x)dx = 1$. Since $f(x) = 0$ for $x \leq 1$, it is clear that we only have to consider
$$\int_1^\infty f(x) dx = \int_1^\infty \frac{c}{x^6} dx = c \cdot \int_1^\infty \frac{1}{x^6} dx = c \cdot \left[ \frac{-1}{5x^5} \right]^\infty_1 = c \cdot \lim_{x \to \infty} \left[ \frac{1}{5} - \frac{1}{5x^5} \right] = \frac c 5$$
And since it must be that $\int_{-\infty}^\infty f(x)dx = 1$, we must have that $\frac c 5 = 1$ so $c=5$.  

*We may calculate $E(X) = \int_{-\infty}^\infty xf(x)dx$. Again, since $f(x) = 0$ for $x \leq 1$, we want to calculate
$$\int_1^\infty xf(x) dx = \int_1^\infty x\frac{5}{x^6}dx = 5\int_1^\infty \frac{1}{x^5}dx = 5 \left[ \frac{-1}{4x^4}\right]^\infty_1 = 5 \lim_{x\to\infty} \left[ \frac 1 4 - \frac{1}{4x^4}\right] = \frac 5 4$$

*We calculate $Var(X) = E(X^2) - (E(X))^2$. We know $(E(X))^2 = \left(\frac{5}{4}\right)^2 = \frac {25}{16}$. We need $E(X^2)$, which we may calculate with:
$$\int_1^\infty x^2f(x) dx = 5\int_1^\infty \frac{1}{x^4}dx = 5 \left[ \frac{-1}{3x^3} \right]_1^\infty = 5 \lim_{x \to \infty} \left[ \frac 1 3 - \frac{1}{3x^3} \right] = \frac 5 3$$
So $Var(X) = \frac 5 3 - \frac{25}{16} = \frac{25}{15} - \frac{25}{16} = 25 \left( \frac{1}{15} - \frac{1}{16}\right) = 25 \left( \frac{16}{15\cdot 16} - \frac{15}{16 \cdot 15}\right) = \frac{25}{16 \cdot 15} = \frac 5 {48}$.

Your reasoning and execution are perfectly correct.
