Cumulative distribution function question 
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Normal random variable $X$ and the cdf of $Y=aX+b$ 

I'm given a standard random variable $X$, and $Y = aX + b$:
How can I find the cumulative distribution function for Y as an integral of $f(x)=(\frac{1}{\sqrt{2}\pi\sigma}e^{-\frac{(x-u)^2}{\sigma^2}}$?
I know $F_y(y)=F_x(\frac{y-b}{a})$, but cant figure out where to go from there.
 A: As an integral of the function you have given, you would need $u=b$ and $\sigma=a$.
Proof: As you note 
$$F_Y(y) = F_X\left(\frac{y-b}{a}\right) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{t} e^{-x^2/2} \,dx, $$
where $t = \frac{y-b}{a}$.  Now make the substitution $x = \frac{s-b}{a}$.  Then $dx = ds/a$, so
$$F_Y(y) = \frac{1}{a\sqrt{2\pi}}\int_{-\infty}^{y} e^{-(s-b)^2/2a^2} \,ds. $$
A: If $a>0$ then
$$
\begin{align}
& f_Y(y) = \frac{d}{dy} F_Y(y) = \frac{d}{dy}\Pr(Y\le y) = \frac{d}{dy}\Pr(aX+b\le y) = \frac{d}{dy}\Pr\left(X \le \frac{y-b}{a}\right) \\  \\
& = \frac{d}{dy}F_X\left( \frac{y-b}{a} \right) = f_X\left(\frac{y-b}{a}\right) \cdot \frac{d}{dy} \frac{y-b}{a} = \frac 1 a f_X\left(\frac{y-b}{a}\right) \\  \\  \\
& = \frac 1 a \cdot \frac{1}{\sqrt{2\pi}}\exp\left(\frac{-1}{2} \cdot \left(\frac{y-b}{a}\right)^2\right).
\end{align}
$$
Therefore
$$
F_Y(y) = \frac 1 a \int_{-\infty}^y \frac{1}{\sqrt{2\pi}} \exp\left(\frac{-1}{2} \left(\frac{u-a}{a}\right)^2\right)\; du.
$$
The equivalence between $aX+b\le y$ and $X \le \frac{y-b}{a}$ holds only if $a>0$.
