Normalization of the wavefunction φ(x) = A
Your derivation is mostly correct, but you made a small mistake when evaluating the integral

\begin{equation}
\int\limits_{-b}^{3b} | \psi (x) |^2 \,\mathrm dx = \int\limits_{-b}^{3b} | A |^2 \,\mathrm dx = 1
\end{equation}

The antiderivative of a constant $C$ is the product of this constant with the integration variable, e.g. $x C$.
So

\begin{align}
\int\limits_{-b}^{3b} \lvert A \rvert^2 \,\mathrm dx = \Bigl[ x \lvert A \rvert^2 \Bigr]_{-b}^{3b} &amp;= \lvert A \rvert^2 \bigl((3b - (-b)\bigr) = 4 b \lvert A \rvert^2 = 1 \\
&amp;\Rightarrow  A = \pm \frac{1}{2 \sqrt{b}}
\end{align}

As for the $\pm$ sign ambiguity: In quantum mechanics, we know that a wave function can always be multiplied by a constant phase factor:

\begin{equation}
\psi'=\mathrm e^{-\mathrm i\alpha}\psi,
\end{equation}

where $\alpha \neq \alpha(x,t)$, because the probability distribution of a system is unaffected by the above transformation, and also the Schrodinger equation and the probability current are unaffected by the above transformation.
So - since $-1 = \mathrm e^{\mathrm{i} \pi}$ and $+1 = \mathrm e^{\mathrm{i} 2 \pi}$ are examples of such constant phase factors - it doesn't make a difference, from a physics point of view, whether you use $+\frac{1}{2 \sqrt{b}}$ or $-\frac{1}{2 \sqrt{b}}$ since both lead to the exact same physical results and are thus physically equivalent.