So nowhere in my textbook does it show how to determine the value of c for any probability density function f(x).
However, I just want to check my work to see if I got it right.
We know that for any pdf f(x) the Integral from infinity to negative infinity = 1, that is:
${\int_{-\infty}^{\infty} f(x)dx\ =} 1 $
However, since we are given two boundaries, 1 and c,
${\int_{1}^{c} f(x)dx\ =} 1 $
since, for all x between 1 and c has the probability density function x.
The definite integral becomes:
$ {1/2c^2 - 1/2 = 1} $
Thus,
$ 1/2c^2 = 1 + 1/2 $
$ c^2 = 3 $
$ c = \sqrt{3} $
Am i correct?
You're correct that
$$\int_{-\infty}^{\infty} f(x)dx = 1 \tag{1}\label{eq1A}$$
Since $f(x) = 0$ for $x \le 0$ and $x \gt c$, you need to split the integral into $2$ parts where it's non-zero, i.e.,
$$\begin{equation}\begin{aligned} 1 & = \int_{0}^{1} x^2dx + \int_{1}^{c} xdx \\ & = \left. \frac{x^3}{3}\right\rvert_{0}^{1} + \left. \frac{x^2}{2}\right\rvert_{1}^{c} \\ & = \frac{1}{3} + \frac{c^2}{2} - \frac{1}{2} \\ & = \frac{c^2}{2} - \frac{1}{6} \end{aligned}\end{equation}\tag{2}\label{eq2A}$$
Note you can't ignore the part where $0 \lt x \lt 1$ as you did in your question. This leads to
$$\begin{equation}\begin{aligned} 1 & = \frac{c^2}{2} - \frac{1}{6} \\ \frac{7}{6} & = \frac{c^2}{2} \\ c^2 & = \frac{7}{3} \\ c & = \sqrt{\frac{7}{3}} \\ c & \approx 1.53 \end{aligned}\end{equation}\tag{3}\label{eq3A}$$
I've shown the result accurate to $2$ decimal places, as requested.