Let's solve the given problems step by step.

Problem Statement:

We are given three probability density functions (PDFs) $f(x)$ and asked to find the cumulative distribution function (CDF) $F(x)$ associated with each, and sketch their graphs.

The CDF $F(x)$ is defined as: $F(x) = \int_{-\infty}^{x} f(t) \, dt$

We will find the CDF for each part.

(a) $f(x) = 3(1 - x)^2$, for $0 < x < 1$, zero elsewhere.

1. CDF: $F(x) = \int_{0}^{x} 3(1 - t)^2 \, dt$ Let's compute the integral.

   $\int 3(1 - t)^2 \, dt = 3 \int (1 - 2t + t^2) \, dt = 3 \left( t - t^2 + \frac{t^3}{3} \right)$ Now, evaluate from 0 to $x$: $F(x) = 3 \left( x - x^2 + \frac{x^3}{3} \right) - 3(0) = 3x - 3x^2 + x^3$ Therefore, the CDF is: $F(x) = 3x - 3x^2 + x^3 \quad \text{for} \, 0 < x < 1$ For $x \leq 0$, $F(x) = 0$ (since $f(x) = 0$). For $x \geq 1$, $F(x) = 1$ (since it's the total probability).

(b) $f(x) = \frac{1}{x^2}$, for $1 < x < \infty$, zero elsewhere.

1. CDF: $F(x) = \int_{1}^{x} \frac{1}{t^2} \, dt$ Let's compute the integral.

   $\int \frac{1}{t^2} \, dt = -\frac{1}{t}$ Now, evaluate from 1 to $x$: $F(x) = -\frac{1}{x} + 1$ Therefore, the CDF is: $F(x) = 1 - \frac{1}{x} \quad \text{for} \, x > 1$ For $x \leq 1$, $F(x) = 0$ (since $f(x) = 0$).

(c) $f(x) = \frac{1}{3}$, for $0 < x < 1$ or $2 < x < 4$, zero elsewhere.

1. CDF:

   We need to handle this in two intervals:

   • For $0 < x < 1$: $F(x) = \int_0^x \frac{1}{3} \, dt = \frac{x}{3}$
   • For $1 \leq x < 2$, $f(x) = 0$, so $F(x) = \frac{1}{3}$ (since no probability is added).
   • For $2 < x < 4$: $F(x) = \frac{1}{3} + \int_2^x \frac{1}{3} \, dt = \frac{1}{3} + \frac{x - 2}{3}$ Simplifying this: $F(x) = \frac{x + 1}{3}$

Therefore, the CDF for this case is: