Q.A continuous random variable $X$ has p.d.f. $f(x)=3x^2$ for $0\le x\le 1$ and $f(x)=0$ otherwise.

Check the total area is $1$, then integrate $x f(x)$ for the mean and $f$ over the tail for the probability.

1. Validity. $f(x)=3x^2\ge0$ on $[0,1]$, and

   $\int_{0}^{1}3x^2\,dx=\big[x^3\big]_{0}^{1}=1-0=1,$

   so it is a valid density.
2. Mean.

   $E(X)=\int_{0}^{1}x\cdot 3x^2\,dx=\int_{0}^{1}3x^3\,dx=\left[\frac{3x^4}{4}\right]_{0}^{1}=\frac{3}{4}.$

   (iii) $P\!\left(X>\tfrac12\right)$ is the area under $f$ from $\tfrac12$ to $1$: $P\!\left(X>\tfrac12\right)=\int_{1/2}^{1}3x^2\,dx=\big[x^3\big]_{1/2}^{1}=1-\left(\frac12\right)^3=1-\frac18=\frac78.$ …

For $f(x)=3x^2$ the cumulative area from $0$ to $t$ is simply $t^3$, so any interval prob …

Computing $E(X)$ as $\int f(x)\,dx$ (that gives $1$) instead of $\int x f(x)\,dx$. Also, for $P(X>\tfrac12)$, integrating from …