Problem 16 - Calculus Early Transcendentals

Question

Sketch the following regions (if a figure is not given) and then find the area. The regions between \(y=\sin x\) and \(y=\sin 2 x,\) for \(0 \leq x \leq \pi\)

Step-by-Step Solution

Answer

Answer: The area between the two graphs is \(\frac{3\sqrt{3}}{4}\).

Step 1: Sketch the Graphs

First, sketch the graphs of \(y = \sin x\) and \(y = \sin 2x\) in the interval \(0 \leq x \leq \pi\). The graph of \(y = \sin x\) has one full oscillation between \(0\) and \(\pi\), while the graph of \(y = \sin 2x\) oscillates twice in the same interval.

Step 2: Identify Intersection Points

To find the intersection points, solve for \(x\) when \(\sin x = \sin 2x\) in the interval \(0 \leq x \leq \pi\): $$ \sin x = \sin 2x \\ x = 0, \frac{\pi}{3}, \frac{2\pi}{3}, \pi $$ There are four intersection points: \((0, 0)\), \(\left(\frac{\pi}{3}, \frac{\sqrt{3}}{2}\right)\), \(\left(\frac{2\pi}{3}, -\frac{\sqrt{3}}{2}\right)\), and \((\pi, 0)\).

Step 3: Set Up the Integral

To set up the integral, we need to analyze the regions bound by both graphs. From \(0\) to \(\frac{\pi}{3}\) and from \(\frac{2\pi}{3}\) to \(\pi\), \(y = \sin x\) is above \(y = \sin 2x\). From \(\frac{\pi}{3}\) to \(\frac{2\pi}{3}\), \(y = \sin x\) is below \(y = \sin 2x\). Therefore, we set up the integral as: $$ Area =\int_{0}^{\frac{\pi}{3}} (\sin x - \sin 2x) dx + \int_{\frac{\pi}{3}}^{\frac{2\pi}{3}} (\sin 2x - \sin x) dx + \int_{\frac{2\pi}{3}}^{\pi} (\sin x - \sin 2x) dx $$

Step 4: Solve the Integral

Now, we solve the integral: $$ Area = \left[-\cos x - \frac{1}{2}\cos 2x\right]_0^{\frac{\pi}{3}} + \left[-\frac{1}{2}\cos 2x - (-\cos x)\right]_\frac{\pi}{3}^{\frac{2\pi}{3}} + \left[-\cos x - \frac{1}{2}\cos 2x\right]_\frac{2\pi}{3}^{\pi} \\ Area = \frac{3\sqrt{3}}{4} $$ The area between the graphs of \(y = \sin x\) and \(y = \sin 2x\) in the interval \(0 \leq x \leq \pi\) is \(\frac{3\sqrt{3}}{4}\).

Key Concepts

Trigonometric Functions

Trigonometric Functions play a crucial role in the mathematical world, especially when analyzing periodic phenomena. In the given exercise, two trigonometric functions are considered: \( y = \sin x \) and \( y = \sin 2x \). These functions are examples of sine functions that oscillate over the x-axis, each with a distinct frequency.

• The function \( y = \sin x \) completes one cycle as \( x \) goes from 0 to \( \pi \), showing one complete wave.


• Meanwhile, the function \( y = \sin 2x \) has a frequency twice that of \( \sin x \), hence it completes two full cycles in the same interval.

Understanding the behavior of these curves not only helps in sketching them accurately but also forms the foundation for more complex functions. When dealing with integration problems like this, observing their intersections and deviations becomes essential in determining areas and understanding the competitive dominance of each function over different intervals.

Area Between Curves

The area between curves is a measure of the space enclosed between two functions over a specific interval. In this exercise, the area between \( y = \sin x \) and \( y = \sin 2x \) in the interval \( 0 \leq x \leq \pi \) is sought. Finding this area is a common application of definite integrals and involves a series of steps.

• First, determine where the two curves intersect within the given interval. This divides the range into sub-intervals.
• These sub-intervals are where one function is above the other, or vice versa.
• The area is computed by integrating the difference between the two functions, from top to bottom, across the intervals. For this problem, it involved three separate integrals, contingent on which curve is above the other.

Integrating these differences gives the total accumulated "height" between the curves, which, in geometrical terms, provides the area. This approach might seem complex, but it is a structured way to apply definite integrals to solve real-world problems.

Intersection Points

Intersection Points between two curves signify where the output or the heights of the functions are identical. These points are vital as they mark boundaries between sub-regions over which functions change their relative positions with respect to each other.

• For the given functions \( y = \sin x \) and \( y = \sin 2x \), the intersection points occur when both give the same y-value for corresponding x-values.
• Solving \( \sin x = \sin 2x \) within the interval \( 0 \leq x \leq \pi \), yields the points: \( x = 0, \frac{\pi}{3}, \frac{2\pi}{3}, \) and \( \pi \).

These intersection x-values delineate sections within the interval. Each section either features \( \sin x \) as the upper curve or \( \sin 2x \). This division is crucial for setting up the definite integrals necessary to find the enclosed area. Recognizing intersection points is particularly useful in various calculus applications, ensuring that any computations involving changes in a function's dominance are accurate.