Discussion
A function \(y(x)\) is defined in the interval [0, 1] on the \(x\)-axis as
\[ y(x) = \begin{cases} 2 & \text{if } 0 \le x \lt \frac{1}{3} \\ 3 & \text{if } \frac{1}{3} \le x \lt \frac{3}{4} \\ 1 & \text{if } \frac{3}{4} \le x \le 1 \end{cases} \]
Which one of the following is the area under the curve for the interval [0, 1] on the \(x\)-axis?
Answer: C
The area is \(\frac{13}{6}\).
The function is a step function, so the area is the sum of three rectangles, each of height \(y\) and width equal to the length of its interval.
\[ \begin{aligned} \text{area} &= 2 \cdot \tfrac{1}{3} + 3 \cdot \left(\tfrac{3}{4} - \tfrac{1}{3}\right) + 1 \cdot \tfrac{1}{4} \\ &= \tfrac{2}{3} + 3 \cdot \tfrac{5}{12} + \tfrac{1}{4} \\ &= \tfrac{8}{12} + \tfrac{15}{12} + \tfrac{3}{12} = \tfrac{26}{12} = \tfrac{13}{6} \end{aligned} \]
The function never drops below 1 on [0, 1], so the area must exceed 1; that rules out \(\frac{5}{6}\) and \(\frac{6}{13}\) at once. \(\frac{6}{5}\) is too small as well, since the first two pieces alone give \(\frac{2}{3} + \frac{5}{4} \gt 1.9\).
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