Question

Find the area of the region trapped between \(y=\sin x\) and \(y=\frac{1}{2}, 0 \leq x \leq 17 \pi / 6\).

Short answer

The area is calculated using integrals of sin functions over defined intervals between points of intersection.

Step-by-step solution

Identify the Functions and Limits

The functions given are \(y = \sin x\) and \(y = \frac{1}{2}\). The region of interest is between \(x = 0\) and \(x = \frac{17\pi}{6}\).

Find Points of Intersection

Set \(\sin x = \frac{1}{2}\). The solutions to this equation within the given interval are the intersection points. The general solutions for \(\sin x = \frac{1}{2}\) are \(x = \frac{\pi}{6} + 2k\pi\) and \(x = \frac{5\pi}{6} + 2k\pi\) where \(k\) is an integer. Substituting \(k\) values within \([0, \frac{17\pi}{6}]\) determines the intersection points: \(x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{13\pi}{6}, \frac{17\pi}{6}\).

Set Up the Integral

The area \(A\) between the curves is given by integrating the difference \(|\sin x - \frac{1}{2}|\) over the segments between the intersection points. The total area is thus split into intervals: \([\frac{\pi}{6}, \frac{5\pi}{6}]\) and \([\frac{13\pi}{6}, \frac{17\pi}{6}]\).

Evaluate Each Integral

Each integral should be evaluated separately. For \([\frac{\pi}{6}, \frac{5\pi}{6}]\), calculate \(\int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} (\sin x - \frac{1}{2}) \, dx\). For \([\frac{13\pi}{6}, \frac{17\pi}{6}]\), calculate \(\int_{\frac{13\pi}{6}}^{\frac{17\pi}{6}} (\sin x - \frac{1}{2}) \, dx\).

Integrate

Find the integral of \(\sin x - \frac{1}{2}\). The integral of \(\sin x\) is \(-\cos x\), and the integral of \(\frac{1}{2}\) is \(\frac{1}{2}x\). Plug in the limits for each integral to find the net area contributions from each section.

Final Area Calculation

For \([\frac{\pi}{6}, \frac{5\pi}{6}]\): \[-\cos x - \frac{x}{2} \bigg|_{\frac{\pi}{6}}^{\frac{5\pi}{6}} = \left(-\cos \frac{5\pi}{6} - \frac{5\pi}{12}\right) - \left(-\cos \frac{\pi}{6} - \frac{\pi}{12}\right) = \frac{1}{4} - \cos \frac{\pi}{6} \]. For \([\frac{13\pi}{6}, \frac{17\pi}{6}]\), repeat the same process. Sum the absolute values of both integrals for the total area.

Key concepts

Trigonometric Functions

Trigonometric functions are essential when dealing with periodic behavior in mathematics, especially functions like sine, cosine, and tangent. In this exercise, the function we are focusing on is the sine function, denoted as \(y = \sin x\). The sine function is periodic with a period of \(2\pi\), repeating its values in regular intervals over this period. This characteristic makes it particularly useful when analyzing wave-like patterns or circular motion.

The sine function ranges from -1 to 1, achieving its maximum at \(\frac{\pi}{2} + 2k\pi\) and its minimum at \(\frac{3\pi}{2} + 2k\pi\), where \(k\) is an integer. This periodic nature and symmetry around the origin are crucial when determining where two curves intersect, which is the first step in finding the area between them. In this exercise, we're comparing \(\sin x\) with the constant function \(y = \frac{1}{2}\), a horizontal line on the y-axis. Understanding these functions' behavior over the specified interval helps us find where they converge, forming the boundaries for integration.

Definite Integral

The definite integral is a fundamental concept in calculus used to calculate the area under a curve over a specific interval. To find the area between two curves, we often calculate the integral of their difference over given intervals. In this problem, we are looking to find the regions where the curves \(y = \sin x\) and \(y = \frac{1}{2}\) diverge, capturing the enclosed area between them.

In general terms, the definite integral is expressed as \(\int_{a}^{b} f(x) \, dx\), where \(a\) and \(b\) are the limits of integration, and \(f(x)\) is the function we are integrating. Here, we are integrating \(\sin x - \frac{1}{2}\), but we take into account absolute values to ensure the area is positive, calculating \(\int (|\sin x - \frac{1}{2}| ) \, dx\). These integrals are evaluated over the segments derived from the intersection points. Integrating in parts and evaluating the limits helps us find the accumulation of areas that lie between the two curves.

Intersection Points

Finding intersection points is crucial when determining the area between two functions. It's the points where two graphs meet or cross over on the coordinate plane. For the functions \(y = \sin x\) and \(y = \frac{1}{2}\), the intersection points are found by setting the two equations equal: \(\sin x = \frac{1}{2}\). Solving this equation will give us multiple intersection points due to the periodic nature of the sine function.

The general solutions to the equation \(\sin x = \frac{1}{2}\) are \(x = \frac{\pi}{6} + 2k\pi\) and \(x = \frac{5\pi}{6} + 2k\pi\) for any integer \(k\). By calculating within a specified domain, here the interval \([0, \frac{17\pi}{6}]\), we identify specific \(k\) values that lead to the intersection points \(x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{13\pi}{6}, \frac{17\pi}{6}\). These points effectively divide the interval into smaller partitions that are used for integrating and finding the area between the curves.

Intervals of Integration

Intervals of integration are segments of the x-axis within which we calculate the area under the curve. In the context of finding the area between two curves, these intervals are determined by the intersection points where the two curves meet. The given problem involves the interval \([0, \frac{17\pi}{6}]\), subdividing it into smaller intervals based on where the graphs intersect, which are \([\frac{\pi}{6}, \frac{5\pi}{6}]\) and \([\frac{13\pi}{6}, \frac{17\pi}{6}]\).

Evaluating the definite integrals across these intervals ensures that the area is accurately calculated across all affected sections. By performing the integration on each interval, and taking absolute values as necessary to ensure positive results, the total area captured between the curves over the entire domain is achieved. This method effectively manages periodic functions, which may cross over each other multiple times within a given range, capturing distinct "trapped" areas that are sought in such problems.