If the probability density function is equal to $mx+b$ for $0<x<5$, and $0$ other wise, what is the probability of $x=1$?

Question

The pdf is $$ f(x) = \left\{ \begin{array}{ll} mx+b & \quad 0<x<5 \\ 0 & \quad \text {otherwise} \end{array} \right. $$

The question is:

What is P(X=1)?

In my textbook, it says $P(X=1)=F(1)-F(1^-)$. F is the cumulative density function CDF.

My answer:$$P(X=1)=\int^1_0 (mx+b) dx-\int^{1^-}_0 (mx+b) dx$$

Is this zero?

Answer 1

If $X$ is a continuous random variable, then by definition $$P(X=1)=0$$

In general, the probability that a continuous random variable takes a specific value $a$ is $0$, because

$$P(X=a) = \int_{\{a\}}f_X\,d\lambda = 0$$

Answer 2

The probability of a continuous variable taking a particular value is 0.

Why is the probability of a continuous variable taking a particular value zero? Explain only logically