normal distribution derivation

Question

In this derivation:

http://www.sonoma.edu/users/w/wilsonst/Papers/Normal/default.html

how do these equal? $$ -k\int (x-\mu) dx = -\frac{k}{2} (x-\mu)^2$$

Isn't this the case?

$$ -k\int (x-\mu) dx = -\frac{kx^2}{2} + k\mu x$$

Answer 1

The answer given comes from one antiderivative of $x-\mu$. Your answer comes from another antiderivative of $x-\mu$. The two differ by a constant, so both of them are correct antiderivatives. Neither of them is the general antiderivative of $x-\mu$.

The general antiderivative of $x-\mu$ can be written as $\frac{1}{2}(x-\mu)^2+C$ or as $\frac{x^2}{2}-\mu x+C$, where in each case the $C$ is an arbitrary constant of integration. Your source presumably found the first form more convenient.