Evaluate general normal distribution $N(\mu, \sigma^2)$ by scaling to $N(0,1)$. Why does integrals result in same value? Suppose $X \sim N(\mu, \sigma^2)$ is a normal RV, $\sigma \neq 0$.
To calculate $P(a < X < b)$ we must compute $$\int^a_b \frac {1} {\sigma \sqrt {2 \pi}} e^{-\frac {(x-\mu)^2} {2\sigma^2}} \, dx$$
However we cannot easily compute this value, so we scale $X$ resulting in a new RV $Z = \frac {X - \mu} {\sigma} \sim N(0,1)$.
Now to compute $P(a < X < b)$ we simply evaluate $$\int^{\frac {a-\mu} {\sigma}}_{\frac {b-\mu} {\sigma}} \frac {1} { \sqrt {2 \pi}} e^{-\frac {x^2} {2}} \, dx$$, which can be done in a table for the standard normal distribution.
However can someone tell me why these two integrals result in the same value ?
Also, to evaluate $P(a < X < b)$ by $$\int^a_b \frac {1} {\sigma \sqrt {2 \pi}} e^{-\frac {(x-\mu)^2} {2\sigma^2}} \, dx$$ are there other methods than scaling to a standard normal variable $Z$ ?
 A: $$
P = \frac {1} {\sigma \sqrt {2 \pi}}\int^a_b  e^{-\frac {(x-\mu)^2} {2\sigma^2}}dx
$$
Forgetting what you know about probability, let's use calculus to change variables in the above integral. Let 
$$z=\frac{x-\mu}{\sigma},$$
then $dz=\frac{dx}{\sigma}$. For the limts, when $x=b$, $z=\frac{b-\mu}{\sigma}$. When $x=a$, $z=\frac{a-\mu}{\sigma}$. Finally, the argument of the exponential, in terms of $z$, is $-\frac{z^2}{2}$Substituting all of that into the integral to change variables, we get
$$
P = \frac {1}{\sqrt {2 \pi}}\int^{\frac{a-\mu}{\sigma}}_{\frac{b-\mu}{\sigma}}  e^{-\frac {z^2} {2}}dz
$$
So, changing variables allows us to kind of standardize things, instead of having a single table for every combination of $\mu,\sigma$, which would be quite a lot of pages of tables!
As for your other question about evaluating the integral directly, it has a closed form expression in terms of error functions. These are tabulated in books as well, and modern calculators often have implementations of the values as well. You could just as easily calculate the integral by numerical quadratures, to any precision you desire. This is likely how the error function or z-tables were generated to begin with.
A: Just do a routine substitution (and don't confuse the letter $u$ with the letter $\mu$):
\begin{align}
u & = \frac{x-\mu}{\sigma} \\[10pt]
du & = \frac{dx}{\sigma}
\end{align}
As $x$ goes from $a$ to $b$, then $u$ goes from $\dfrac{a-\mu}{\sigma}$ to $\dfrac{b-\mu}{\sigma}$.
