Q.The probability density function of $X$ is given by $f(x)=\begin{cases}kxe^{-2x} & x>0\\ 0 & x\le0\end{cases}$. Find the value of $k$.

Cumulative distribution function (both cases). For any random variable $X$, the cdf $F(x)=P(X\le x)$ is defined for every real $x$.

• Discrete (Definition 11.4): $F(x)=\sum_{x_i\le x}f(x_i)$ — a step function, constant between support points and jumping by $f(x_i)$ at each $x_i$. Conversion both ways: given the pmf, $F$ is the running (cumulative) sum of $f$ up to $x$; given $F$, the pmf is recovered as the jump size $f(x_i)=F(x_i)-F(x_{i-1})$ at each point of discontinuity (with $F(x_0)=0$ before the first jump) — the jump of $F$ at $a$ is exactly $P(X=a)$.
• Continuous (Definition 11.7): $F(x)=\int_{-\infty}^x f(u)\,du$ — here $F$ is everywhere continuous (no jumps, since no single point carries probability). Conversion both ways: given the pdf, integrate piece by piece to build $F$; given $F$, differentiate — $f(x)=F'(x)$ wherever the derivative exists (at the finitely many "corner" points, $f$ may be set to any convenient value, since it never affects an interval probability).

Probability density function (Definition 11.6): a non-negative $f(x)$ is a pdf if $P(a\le X\le b)=\int_a^b f(x)\,dx$ for every $a\le b$. By Theorem 11.2, $f$ is a valid pdf exactly when (i) $f(x)\ge0$ everywhere, and (ii) $\int_{-\infty}^\infty f(x)\,dx=1$ (total area $=1$) — the continuous analogue of Theorem 11.1, used the same way to solve for an unknown normalising constant. Because $P(X=a)=\int_a^a f=0$ always, the four inequality forms $P(a\le X\le b)=P(a<X\le b)=P(a\le X<b)=P(a<X<b)$ all coincide for a continuous $X$.

Standing cdf properties. Both cases share: $0\le F(x)\le1$; $F$ non-decreasing; $\lim_{x\to-\infty}F(x)=0$, $\lim_{x\to\infty}F(x)=1$; and the interval formula $P(a\le X\le b)=F(b)-F(a)$ (continuous case) or $P(x_1<X\le x_2)=F(x_2)-F(x_1)$ (discrete case) — letting one avoid re-integrating/re-summing for every new interval once $F$ is known.