Suppose that the p.d.f. of a random variable $X$ is as follows: $$ f(x) = \ \begin{cases} \dfrac{c}{(1−x)^{1/2}} & \text{for } 0 < x < 1 \\[10pt] 0 & \text{otherwise} \end{cases} \ $$
Solution: $ \displaystyle \int_0^1 \frac{c}{(1−x)^{1/2}} \, dx=2c=1$, $c=\dfrac{1}{2}$
I don't understand how $c$ is $\frac{1}{2}$ and not $\frac{-1}{2}$ because in integrating it should be $-2c$ and not $2c$. Is this a property of p.d.f?
$$ \int_0^1 \frac{dx}{\sqrt{1-x}} $$
In this integral, note that if $0<x<1$ then $\dfrac 1 {\sqrt{1-x}}$ is positive. And it's $\displaystyle\int_0^1,$ not $\displaystyle \int_1^0.$ Therefore the value of the integral must be positive.
\begin{align} u & = \sqrt{1-x} \\ u^2 & = 1-x \\ 2u\,du & = -dx \\ \text{As } x & \text{ goes from $0$ to $1$, $u$ goes from $1$ to $0$.} \\[10pt] \int_0^1 \frac{dx}{\sqrt{1-x}} & = \int_1^0 \frac{-du}{\sqrt u} = \left. -2\sqrt u \vphantom{\frac 1 1}~\right|_1^0 = -2(0-1) = 2. \end{align}