Area Between Curves -sin(2x) And -sin(4x) On [-π/8, Π/8]

by ADMIN 57 views

Finding the area between two curves is a fundamental concept in calculus, with wide-ranging applications in physics, engineering, and economics. This article provides a detailed walkthrough of how to calculate the area of the region bounded by the curves f(x) = -sin(2x) and g(x) = -sin(4x) over the interval [-π/8, π/8]. We will explore the necessary steps, from identifying intersection points to setting up and evaluating the definite integral. Understanding these techniques is crucial for anyone studying calculus or related fields. So, let’s dive in and unravel the solution step by step.

Understanding the Problem

Before we begin the calculations, it's essential to fully grasp the problem at hand. The task is to find the area A of the region nestled between two trigonometric curves: f(x) = -sin(2x) and g(x) = -sin(4x). However, we're not looking for the area over the entire real number line; instead, we're focusing on a specific interval, [-π/8, π/8]. This interval acts as our boundaries, defining the left and right edges of the region we're interested in. The functions f(x) and g(x), both sine functions with different frequencies, will intersect and create enclosed spaces. Our goal is to determine the total area of these spaces within the given interval. To achieve this, we need to identify where these curves intersect, which curve lies above the other within different subintervals, and finally, set up and evaluate the appropriate definite integrals.

Step 1: Finding the Intersection Points

The first crucial step in finding the area between curves is to identify the points of intersection. These points mark where the curves f(x) and g(x) meet, which is where f(x) = g(x). In our case, this translates to solving the equation -sin(2x) = -sin(4x). To solve this trigonometric equation, we can start by multiplying both sides by -1, which simplifies the equation to sin(2x) = sin(4x). Next, we can employ the double-angle identity for sine, which states that sin(2θ) = 2sin(θ)cos(θ). Applying this to the right side of our equation, we get sin(2x) = 2sin(2x)cos(2x). Rearranging the terms gives us 2sin(2x)cos(2x) - sin(2x) = 0. We can factor out sin(2x), leading to sin(2x)(2cos(2x) - 1) = 0. This equation holds true if either sin(2x) = 0 or 2cos(2x) - 1 = 0. Let's solve each case separately.

For sin(2x) = 0, the solutions occur when 2x = nπ, where n is an integer. Dividing by 2, we get x = nπ/2. Within our interval [-π/8, π/8], the only solution that falls within this range is x = 0. For the second case, 2cos(2x) - 1 = 0, we can rearrange it to cos(2x) = 1/2. The solutions for this equation occur when 2x = ±π/3 + 2nπ, where n is an integer. Dividing by 2, we have x = ±π/6 + nπ. Within our interval [-π/8, π/8], the solutions are x = -π/6 and x = π/6. Therefore, the intersection points within the given interval are x = -π/6, x = 0, and x = π/6. These points are vital because they divide the interval into subintervals where one function consistently lies above the other. Knowing these intersection points is crucial for setting up the integrals correctly to calculate the area.

Step 2: Determining the Intervals and Which Function is Greater

With the intersection points identified, the next step is to determine the intervals created by these points within the given range and to ascertain which function is greater within each interval. Our interval [-π/8, π/8] is now divided into subintervals by the intersection points x = -π/6, x = 0, and x = π/6. This gives us three subintervals to consider: [-π/8, -π/6], [-π/6, 0], and [0, π/6]. Notice that π/6 is greater than π/8, so the subinterval [-π/8, -π/6] should not exist. There are only 2 subintervals, [-π/8, 0], and [0, π/8]. For each of these intervals, we need to figure out whether f(x) = -sin(2x) is greater than g(x) = -sin(4x), or vice versa.

To do this, we can select a test point within each interval and evaluate both functions at that point. The function with the larger value at the test point is the greater function within that interval. Let's start with the interval [-π/8, 0]. A convenient test point within this interval is x = -π/12. Evaluating f(-π/12) = -sin(2(-π/12)) = -sin(-π/6) = 1/2. Evaluating g(-π/12) = -sin(4(-π/12)) = -sin(-π/3) = √3/2. Since √3/2 > 1/2, we conclude that g(x) is greater than f(x) on the interval [-π/8, 0]. Now, let's consider the interval [0, π/8]. A suitable test point here is x = π/12. Evaluating f(π/12) = -sin(2(π/12)) = -sin(π/6) = -1/2. Evaluating g(π/12) = -sin(4(π/12)) = -sin(π/3) = -√3/2. In this case, -1/2 > -√3/2, so f(x) is greater than g(x) on the interval [0, π/8]. This determination is crucial for setting up the correct integral for calculating the area, as we need to subtract the smaller function from the larger function within each interval.

Step 3: Setting Up the Integrals

Now that we've identified the intervals and determined which function is greater on each interval, we're ready to set up the integrals that will calculate the area. The area A between the curves f(x) and g(x) over the interval [-π/8, π/8] can be found by summing the areas of the regions formed in each subinterval. Remember, the area between two curves is given by the integral of the absolute difference between the functions. This ensures that we're always integrating a positive quantity, as area cannot be negative.

In our case, we have two subintervals: [-π/8, 0] and [0, π/8]. On the interval [-π/8, 0], we found that g(x) = -sin(4x) is greater than f(x) = -sin(2x). Therefore, the area in this interval is given by the integral ∫[-π/8 to 0] (g(x) - f(x)) dx = ∫[-π/8 to 0] (-sin(4x) - (-sin(2x))) dx = ∫[-π/8 to 0] (-sin(4x) + sin(2x)) dx. On the interval [0, π/8], we found that f(x) = -sin(2x) is greater than g(x) = -sin(4x). Thus, the area in this interval is given by the integral ∫[0 to π/8] (f(x) - g(x)) dx = ∫[0 to π/8] (-sin(2x) - (-sin(4x))) dx = ∫[0 to π/8] (-sin(2x) + sin(4x)) dx. The total area A is the sum of these two integrals: A = ∫[-π/8 to 0] (-sin(4x) + sin(2x)) dx + ∫[0 to π/8] (-sin(2x) + sin(4x)) dx. This setup is crucial; it breaks down the problem into manageable parts that we can evaluate using the fundamental theorem of calculus.

Step 4: Evaluating the Integrals

With the integrals set up, the next step is to evaluate them. We have two integrals to compute: ∫[-π/8 to 0] (-sin(4x) + sin(2x)) dx and ∫[0 to π/8] (-sin(2x) + sin(4x)) dx. Let's evaluate them separately.

For the first integral, ∫[-π/8 to 0] (-sin(4x) + sin(2x)) dx, we need to find the antiderivative of -sin(4x) + sin(2x). The antiderivative of -sin(4x) is (1/4)cos(4x), and the antiderivative of sin(2x) is -(1/2)cos(2x). So, the antiderivative of -sin(4x) + sin(2x) is (1/4)cos(4x) - (1/2)cos(2x). Now we evaluate this antiderivative at the limits of integration: [(1/4)cos(4(0)) - (1/2)cos(2(0))] - [(1/4)cos(4(-π/8)) - (1/2)cos(2(-π/8))] = [(1/4)cos(0) - (1/2)cos(0)] - [(1/4)cos(-π/2) - (1/2)cos(-π/4)] = [(1/4)(1) - (1/2)(1)] - [(1/4)(0) - (1/2)(√2/2)] = [1/4 - 1/2] - [0 - √2/4] = -1/4 + √2/4.

Now let's evaluate the second integral, ∫[0 to π/8] (-sin(2x) + sin(4x)) dx. The antiderivative of -sin(2x) + sin(4x) is (1/2)cos(2x) - (1/4)cos(4x). Evaluating this antiderivative at the limits of integration, we get: [(1/2)cos(2(π/8)) - (1/4)cos(4(π/8))] - [(1/2)cos(2(0)) - (1/4)cos(4(0))] = [(1/2)cos(π/4) - (1/4)cos(π/2)] - [(1/2)cos(0) - (1/4)cos(0)] = [(1/2)(√2/2) - (1/4)(0)] - [(1/2)(1) - (1/4)(1)] = [√2/4 - 0] - [1/2 - 1/4] = √2/4 - 1/4. Finally, we add the results of the two integrals to find the total area: A = (-1/4 + √2/4) + (√2/4 - 1/4) = -1/4 + √2/4 + √2/4 - 1/4 = (2√2)/4 - 2/4 = (√2 - 1)/2. Therefore, the area A of the region bounded between the curves is (√2 - 1)/2 square units.

Final Answer

The area A of the region bounded between the curves f(x) = -sin(2x) and g(x) = -sin(4x) over the interval [-π/8, π/8] is (√2 - 1)/2 square units. This result is obtained by meticulously following the steps of finding intersection points, determining intervals, setting up definite integrals, and evaluating them. This process showcases the power of calculus in solving geometric problems and provides a solid foundation for tackling similar challenges in the future.