Question

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

• A. \( \frac{5}{6} \)
• B. \( \frac{6}{5} \)
• C. \( \frac{13}{6} \)
• D. \( \frac{6}{13} \)

Hint:

When calculating the area under a piecewise function, break it into intervals and calculate the area for each segment separately.

Answer 1

The Correct Option is C

To find the area under the curve of the given piecewise function \( y(x) \) over the interval [0, 1], we must calculate the integral of \( y(x) \) in each sub-interval and sum these areas. The function \(\ y(x) \) is defined as:

\[ y(x) = \begin{cases} 2 & \text{if } 0 \leq x < \frac{1}{3} \\ 3 & \text{if } \frac{1}{3} \leq x < \frac{3}{4} \\ 1 & \text{if } \frac{3}{4} \leq x \leq 1 \end{cases} \]

We will calculate the area for each segment: 

1. Interval: \( [0, \frac{1}{3}] \)
   For this interval, \( y(x) = 2 \). \(Area_1 = \int_{0}^{\frac{1}{3}} 2 \, dx = 2 \left[ x \right]_{0}^{\frac{1}{3}} = 2 \left( \frac{1}{3} - 0 \right) = \frac{2}{3}\)
2. Interval: \( [\frac{1}{3}, \frac{3}{4}] \)
   For this interval, \( y(x) = 3 \). \(Area_2 = \int_{\frac{1}{3}}^{\frac{3}{4}} 3 \, dx = 3 \left[ x \right]_{\frac{1}{3}}^{\frac{3}{4}} = 3 \left( \frac{3}{4} - \frac{1}{3} \right)\)

To solve the expression: \(\frac{3}{4} - \frac{1}{3} = \frac{9}{12} - \frac{4}{12} = \frac{5}{12}\)

1. \(Area_2 = 3 \times \frac{5}{12} = \frac{15}{12} = \frac{5}{4}\)
2. Interval: \( [\frac{3}{4}, 1] \)
   For this interval, \( y(x) = 1 \). \(Area_3 = \int_{\frac{3}{4}}^{1} 1 \, dx = 1 \left[ x \right]_{\frac{3}{4}}^{1} = 1 \left( 1 - \frac{3}{4} \right) = \frac{1}{4}\)

The total area under the curve is the sum of the areas from each sub-interval:

\(\text{Total Area} = Area_1 + Area_2 + Area_3 = \frac{2}{3} + \frac{5}{4} + \frac{1}{4}\)

Convert all fractions to have a common denominator:

\(\frac{2}{3} = \frac{8}{12}, \quad \frac{5}{4} = \frac{15}{12}, \quad \frac{1}{4} = \frac{3}{12}\)

Adding these fractions, we get:

\(\frac{8}{12} + \frac{15}{12} + \frac{3}{12} = \frac{26}{12} = \frac{13}{6}\)

Thus, the area under the curve for the interval [0, 1] is \(\frac{13}{6}\).