Problem 25 - Applied Calculus

Question

Use an integral to find the specified area. Between \(y=\sin x+2\) and \(y=0.5\) for \(6 \leq x \leq 10.\)

Step-by-Step Solution

Answer

The area between the curves from \( x = 6 \) to \( x = 10 \) is approximately 5.88 square units.

Step 1: Setup the Problem

We have two functions, the sinusoidal function: \( y = \sin x + 2 \) and a line: \( y = 0.5 \). We need to find their points of intersection in the interval \( 6 \leq x \leq 10 \).

Step 2: Find Points of Intersection

Set the two functions equal to find the points of intersection: \( \sin x + 2 = 0.5 \). Simplifying gives \( \sin x = -1.5 \), which is impossible as \( \sin x \) ranges from -1 to 1. Hence, there are no intersection points within the interval.

Step 3: Identify Integration Bounds

Since there are no intersection points, determine the values of \( x \) where \( \sin x + 2 \) consistently lies completely above 0.5 in the interval [6, 10]. Calculate approximate values to verify positions of \( \sin x + 2 \). Evaluate if required using additional techniques such as graph illustration or estimation.

Step 4: Integrate the Function

Calculate the definite integral of the difference between the upper function, \( \sin x + 2 \), and the lower function, 0.5, in the interval [6,10].\[ \int_{6}^{10} ((\sin x + 2) - 0.5) \, dx = \int_{6}^{10} (\sin x + 1.5) \, dx \]

Step 5: Solve the Integral

Calculate the integral \[ \int (\sin x + 1.5) \, dx = -\cos x + 1.5x + C \] Evaluate this from \( x = 6 \) to \( x = 10 \).

Step 6: Evaluate Definite Integral

Substitute the limits of integration into the antiderivative:\[ \left. -\cos x + 1.5x \right|_6^{10} = (-\cos 10 + 15) - (-\cos 6 + 9) \]Calculate approximately (need numerical cosine values), if necessary using a calculator.

Step 7: Calculate the Area

Calculate and simplify the expression from the previous step using appropriate values or tools.Assuming \(-\cos 10 \approx 0.839 \) and \(-\cos 6 \approx 0.960\), we find:\[ (0.839 + 15) - (0.960 + 9) = 5.879 \] or approximately 5.88.

Key Concepts

Integral Calculus

Integral calculus plays a vital role in finding areas and quantities where a sum of many small components is involved. One of its most common applications is determining the area between curves or under a curve itself. The concept relies on understanding how to compute definite integrals. In essence, a definite integral calculates the net area between a function and the x-axis over a specified interval. This technique is widely used across sciences and engineering to determine quantities like distance, area, and even probabilities. When dealing with integrals, we often encounter the Fundamental Theorem of Calculus. This theorem simplifies integration by linking the antiderivative, which allows us to find exact areas or values directly. Working through problems, such as the one given, shows us how to apply integral calculus concepts to determine the area between two functions.

Area Between Curves

Finding the area between curves is a crucial aspect of integral calculus. To determine this area, we subtract the lower function from the upper function over a specified range. This action effectively "cuts out" or isolates the space between the two graphs, giving us the precise area.In our exercise, we examine the functions \( y=\sin x+2 \) and \( y=0.5 \) within the domain \([6, 10]\). The lack of intersection points suggests that the sinusoidal function lies entirely above the line \( y=0.5 \) throughout this range. Therefore, calculating the area is straightforward using the definite integral:

• Identify the functions: upper \( y=\sin x+2 \) and lower \( y=0.5 \).
• Determine integration bounds: \( x = 6 \) to \( x = 10 \).
• Set up the integral: \( \int_{6}^{10} ((\sin x + 2) - 0.5) \, dx \).

The area calculation confirms that \( \sin x + 2 \) maintains its position above 0.5, letting you compute the definite integral smoothly.

Trigonometric Functions

Trigonometric functions like sine, cosine, and tangent are key players in calculus. They capture the essence of periodic phenomena, ranging from wave behavior to oscillations. The function \( y=\sin x+2 \) combines the sine wave with a vertical shift, moving the graph upwards by 2 units.This vertical shift is crucial when determining areas. In the given exercise, noticing how \( \sin x + 2 \) behaves within the interval is essential. While \( \sin x \) oscillates between -1 and 1, adding 2 moves its range to 1 and 3. This ensures it stays above the line \( y = 0.5 \) over the given interval, simplifying area calculations:

• Understand that trigonometric functions can be modified by shifts and stretches.
• Observe the changes on the graph due to vertical shifts.
• Use these shifts to identify points or regions effectively.

Realizing the modifications and their impact aids in efficiently solving integral problems like this.

Integration Techniques

Integration techniques offer diverse tools for solving integrals, particularly when functions are complex. Understanding which method to employ is crucial for simplifying and correctly solving integrals. In this exercise, using basic antiderivatives and the properties of definite integrals suffices.Key techniques include:

• Antiderivatives: Find the antiderivative or indefinite integral of a function to use in evaluation.
• Substitution: Occasionally, replacing a variable simplifies the integral.
• Definite Integrals: After finding the antiderivative, evaluate it at the given limits to find the total area or value.

Here, the antiderivative \( -\cos x + 1.5x \) helps find the area by simply evaluating from 6 to 10. This showcases the elegance and power of integration in solving real-world area problems.