Question
Let $X$ be a continuous random variable whose probability density function is $f(x)=3 x^2$, for $0<x<1$. note that $f(x)$ is not $P[X=x]$.
For example, $f(0.9)=3(0.9)^2=2 \cdot 43>1$, which is clearly not a probability. In the continuous case, $f(x)$ is the height of the curve at $X=x$, so that the total area under the curve is 1 . Here it is areas under the curve that define the probabilities.

Answer

Now, let's start by verifying that $f(x)$ is a valid probability density function.
For this, note the following results.
1. $f(x)=3 x^2 \geq 0$ for all $x \in[0,1]$.
2. $\int_0^1 f(x)=\int_0^1 3 x^2 d x=1$
Therefore, the function $f(x)=3 x^2$, for $0<x<1$ is a proper probability density function.

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