What choices of $c, d$ would make $\int_{x=-\infty}^{x_0} (cx - d) \mathcal{N}(x; \mu, \sigma^2)\,dx $ positive? Consider the following integration: 
$$
I(x_0|c, d) = \int_{x=-\infty}^{x_0} (cx - d) \mathcal{N}(x; \mu, \sigma^2) \,dx
$$
In this notation, $\mathcal{N}(x; \mu, \sigma^2)$ is a normal distribution: 
$$
\mathcal{N}(x; \mu, \sigma^2)  = \frac{1}{\sqrt{2\pi\sigma^2} } e^{ -\frac{(x-\mu)^2}{2\sigma^2} }
$$
For a fixed and finite $x_0$ what choices of $d$ and $c$ would result in positive $I$?
 A: So $\mathcal{N}(x; \mu, \sigma^2)$ is the density of a normal distribution. Let $F(x; \mu,\sigma^2)$ be the distribution function for this normal distribution.
First
$$I(x_0\ |\ c,d) = c\int_{x=-\infty}^{x_0} x \mathcal{N}(x; \mu, \sigma^2)dx - d\int_{x=-\infty}^{x_0} \mathcal{N}(x; \mu, \sigma^2)dx.$$
You can check that
$$\frac{d}{dx}\mathcal{N}(x; \mu, \sigma^2) = -\frac{x-\mu}{\sigma^2}\mathcal{N}(x; \mu, \sigma^2).$$
Thus
\begin{align*}
I(x_0\ |\ c,d) &= -c\sigma^2\int_{x=-\infty}^{x_0}-\frac{x-\mu}{\sigma^2} \mathcal{N}(x; \mu, \sigma^2)dx + (c\mu + d)\int_{x=-\infty}^{x_0} \mathcal{N}(x; \mu, \sigma^2)dx \\
&=-c\sigma^2\int_{x=-\infty}^{x_0} \frac{d}{dx}\mathcal{N}(x; \mu, \sigma^2)dx + (c\mu + d)F(x_0;\mu,\sigma^2)\\
&= -c\sigma^2 \mathcal{N}(x_0; \mu, \sigma^2) + (c\mu + d)F(x_0;\mu,\sigma^2).
\end{align*}
Both $\mathcal{N}(x; \mu, \sigma^2)$ and $F(x; \mu,\sigma^2)$ are known. For $I(x_0\ |\ c,d) > 0$ to hold, you want 
$$(c\mu + d)\frac{F(x_0;\mu,\sigma^2)}{\mathcal{N}(x_0; \mu, \sigma^2)} > c\sigma^2.$$
At this point, I would split into a couple of cases. If $c\mu+d>0$, then you want to choose $c, d$ such that 
$$\frac{F(x_0;\mu,\sigma^2)}{\mathcal{N}(x_0; \mu, \sigma^2)} > \frac{c\sigma^2}{(c\mu + d)},$$
and with the reverse inequality if $c\mu+d<0$.
