Q.Find $k$, if $f(x)=kx^2(1-x)$, for $0<x<1$, $=0$ otherwise, is the p.d.f. of random variable $X$.

For a continuous random variable $X$ defined on an interval $S = (a, b)$, a probability density function $f(x)$ is a non-negative, integrable function satisfying three conditions: $f(x) \ge 0$ for every $x$ in $S$; the total area under the curve $y = f(x)$ over $S$ equals 1, i.e. $\int_S f(x)\,dx = 1$; and for any interval $A$ inside $S$, the probability that $X$ falls in $A$ is the area under $f$ over that interval, $P[X \in A] = \int_A f(x)\,dx$. Crucially, $f(x)$ itself is not a probability (it can exceed 1) — it is only areas under the curve, obtained by integrating $f$, that …