A function $y(x)$ is defined in the interval $[0, 1]$ on the $x$-axis as $$y(x) = \begin{cases} 2 & \text{if } 0 \le x < \frac{1}{3} \[6pt] 3 & \text{if } \frac{1}{3} \le x < \frac{3}{4} \[6pt] 1 & \text{if } \frac{3}{4} \le x \le 1 \end{cases}$$ Which one of the following is the area under the curve for the interval $[0, 1]$ on the $x$-axis?

The area under the piecewise continuous curve $y(x)$ on the interval $[0, 1]$ is given by the definite integral:

Evaluating each definite integral as the area of a rectangle ($\text{height} \times \text{width}$):

1. For $0 \le x < \frac{1}{3}$:

   $$\text{Area}_1 = 2 \times \left(\frac{1}{3} - 0\right) = \frac{2}{3}$$

2. For $\frac{1}{3} \le x < \frac{3}{4}$:

   $$\text{Area}_2 = 3 \times \left(\frac{3}{4} - \frac{1}{3}\right) = 3 \times \frac{5}{12} = \frac{5}{4}$$

3. For $\frac{3}{4} \le x \le 1$:

   $$\text{Area}_3 = 1 \times \left(1 - \frac{3}{4}\right) = 1 \times \frac{1}{4} = \frac{1}{4}$$

Summing all three sub-areas:

Hence, the correct option is (C).