Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
The wave function $ \psi(x) $ is given as:
$$
\psi(x) =
\begin{cases}
cx & |x| \leq 1 \; \rm{nm} \\\\
\dfrac{c}{x} & |x| \geq 1 \; \rm{nm}
\end{cases}
$$
To find the normalization constant $ c $, we use the normalization condition, which states that the total probability must equal 1.
$$ \int_{-\infty}^{\infty} |\psi(x)|^2 \; dx = 1\tag 1 $$
This function is defined piecewise for different ranges of $ x $, but the key detail is how the conditions $ |x| \leq 1 $ and $ |x| \geq 1 $ translate into intervals of $ x $.
$\bullet$ For $ |x| \leq 1 $ : The condition $ |x| \leq 1 $ means $ x $ lies in the interval $ -1 \leq x \leq 1 $. So, the interval for this part of the function is $ [-1, 1] $.
$\bullet$ For $ |x| \geq 1 $ : The condition $ |x| \geq 1 $ splits into two separate cases: $ x \geq 1 $ and $ x \leq -1 $.
$\bullet$$\bullet$ For $ x \geq 1 $, the interval is $ [1, \infty) $.
$\bullet$$\bullet$ For $ x \leq -1 $, the interval is $ (-\infty, -1] $.
$\Rightarrow$$\bullet$$\bullet$ This function has 3 intervals:
1. $ (-\infty, -1] $ with $\psi(x)=\dfrac{c}{x} $, and $|\psi(x)|^2=\dfrac{c^2}{x^2}$
2. $ [-1, 1] $ with $ \psi(x)=cx $, and $|\psi(x)|^2=c^2x^2$
3. $ [1, \infty) $ with $\psi(x)= \dfrac{c}{x} $, and $|\psi(x)|^2=\dfrac{c^2}{x^2}$
Plug that into (1);
$$\int_{-\infty}^{-1} \dfrac{c^2}{x^2} \; dx + \int_{-1}^{1} c^2x^2 \; dx+\int_{1}^{\infty} \dfrac{c^2}{x^2} \; dx= 1 $$
Thus,
$$2\int_{ 1}^{ \infty} \dfrac{c^2}{x^2} \; dx +2 \int_{0}^{1} c^2x^2 \; dx= 1 \tag 2$$
Solving for the first integral:
$$ 2 \int_{1}^{\infty} \frac{c^2}{x^2} \; dx = 2c^2 \int_{1}^{\infty} \frac{1}{x^2} \; dx= 2c^2 \left[ -\frac{1}{x} \right]_{1}^{\infty} $$
$$2 \int_{1}^{\infty} \frac{c^2}{x^2} \; dx = 2c^2 \left( 0 + \frac{1}{1} \right) = 2c^2\tag 3 $$
Solving for the second integral:
$$
2 \int_{0}^{1} c^2 x^2 \; dx = 2c^2 \int_{0}^{1} x^2 \; dx= 2c^2 \left[ \frac{x^3}{3} \right]_{0}^{1} $$
$$2 \int_{0}^{1} c^2 x^2 \; dx= 2c^2 \times \frac{1}{3} = \dfrac{2c^2}{3} \tag 4$$
Plug (3) and (4) into (2);
$$2c^2+\dfrac{2c^2}{3}= 1 $$
$$ c^2 = \frac{3}{8}$$
$$ c = \sqrt{\dfrac{3}{8}} = \color{red}{\bf \dfrac{\sqrt{6}}{4}} $$
$$\color{blue}{\bf [b,c]}$$
$$
\psi(x) =
\begin{cases}
\dfrac{c}{x} & -\infty \leq x \leq-1 \\\\
cx &-1 \leq x \leq 1 \\\\
\dfrac{c}{x} & 1 \leq x \leq \infty
\end{cases}
$$
$$
|\psi(x)|^2 =
\begin{cases}
\dfrac{c^2}{x^2} & -\infty \leq x \leq-1 \\\\
c^2x^2 &-1 \leq x \leq 1 \\\\
\dfrac{c^2}{x^2} & 1 \leq x \leq \infty
\end{cases}
$$
See the two graphs below.
$$\color{blue}{\bf [d]}$$
We need to find the probability in the middle interval, so
$$\text{Prob}(-1.0 \, \text{nm} \leq x \leq 1.0 \, \text{nm}) = \int_{-1.0 }^{1.0 } c^2 x^2 dx =\dfrac{N}{N_{tot}}$$
Thus,
$$N=N_{tot} \int_{-1.0 }^{1.0 } c^2 x^2 dx $$
$$N=2c^2N_{tot} \int_{0 }^{1.0 } x^2 dx=\dfrac{ 2c^2N_{tot}}{3}x^3\bigg|_0^1$$
$$N =\dfrac{ 2c^2N_{tot}}{3} (1)^3 $$
Plug the known
$$N =\dfrac{ 2\left[\frac{3}{8}\right](10^6)}{3} $$
$$N =\color{red}{\bf 2.5\times 10^5}\;\rm electron $$
