Problem 43 - Calculus of a Single Variable

Question

Finding the Area of a Region In Exercises \(43-46,(\text { a) use }\) a graphing utility to graph the region bounded by the graphs of the equations, (b) find the area of the region, and (c) use the integration capabilities of the graphing utility to verify your results. $$ f(x)=2 \sin x+\sin 2 x, \quad y=0, \quad 0 \leq x \leq \pi $$

Step-by-Step Solution

Answer

The area of the region is obtained as the definite integral of the function from 0 to \(\pi\). The exact value is obtained by evaluating the antiderivative at the upper and lower limits and subtracting the two results.

Step 1: Graph the Function and Identify the Region

Use a graphing utility to plot the function \(f(x)=2 \sin(x)+\sin(2x)\) for \(0 \leq x \leq \pi\). The region of interest is the area under the curve of the function that is above the x-axis between \(x=0\) and \(x=\pi\).This step should give a rough idea of the shape of the region, which will help in conceptualizing the problem.

Step 2: Set Up and Evaluate the Integral

We will find the exact area of the region by integrating the function from 0 to \(\pi\). In mathematical terms, this is \(\int_{0}^{\pi} (2sin(x) + sin(2x)) dx\). The integral of \(2sin(x)\) is \(-2cos(x)\) and the integral of \(sin(2x)\) is \(-1/2cos(2x)\), so the antiderivative (i.e., the indefinite integral) of \(2sin(x) + sin(2x)\) is -2cos(x) - 1/2cos(2x). Applying the Fundamental Theorem of Calculus, we evaluate this at \(\pi\) and 0 and subtract the two results.

Step 3: Verify the Result

The area under the curve as determined by the integral may be cross-verified utilizing the graphing utility's integration functionality. By comparing thecomputed area to the area outputted by the utility's built-in function, the results may be checked.

Key Concepts

Integration

Integration is a fundamental mathematical concept that involves finding the area under a curve. In our exercise, we seek the area bounded by the function \( f(x)=2 \sin(x)+\sin(2x) \) and the x-axis over the interval \( [0, \pi] \). To find this area, we use integrals.
Let's break down the integration process:

• Set up the integral: Determine the function to integrate and the limits of integration—for this exercise, from 0 to \( \pi \).
• Find the antiderivative: This function, which we find by reversing the differentiation process, lets us evaluate the integral. For our function, the antiderivative is \( -2\cos(x) - \frac{1}{2}\cos(2x) \).
• Apply the Fundamental Theorem of Calculus: Evaluate the antiderivative at the upper limit (\( \pi \)) and subtract the evaluation at the lower limit (0) to find the area.

These steps allow us to compute the exact area of the region bounded by our function.

Graphing utility

Graphing utilities are powerful tools used to visualize functions and the regions under curves. In the context of our exercise, we utilize a graphing utility to plot \( f(x)=2\sin(x)+\sin(2x) \) over the interval \( [0, \pi] \). By doing this, we can see:

• The shape and position of the function relative to the x-axis, which helps in understanding which regions contribute to the area.
• The exact bounds of integration by visually inspecting where the function crosses the x-axis.

After plotting the function, graphing utilities often have built-in capabilities to compute integrals. This feature allows us to verify our manual calculations by comparing the graphing utility's output with our analytically computed area.
Using graphing utilities simplifies the process of solving integration problems by providing both visual confirmation and computational support.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus connects the concepts of derivatives and integrals, providing a way to evaluate definite integrals. In essence, it states that if a function is continuous over an interval, its integral can be determined with the help of an antiderivative.
Here's how it applies to our exercise:

• First, we find the antiderivative of the function \( f(x)=2\sin(x)+\sin(2x) \), which helps us transition from a function representation to an area representation.
• We then evaluate this antiderivative at the upper and lower bounds of our interval, \( \pi \) and 0, giving us the net area under the curve.
• By subtracting these evaluations, we obtain the precise area calculation.

This theorem is a foundational principle in calculus, facilitating the simplification of complex area calculations by connecting them directly to antiderivatives.

Trigonometric functions

Trigonometric functions such as sine and cosine appear regularly in calculus problems, especially when dealing with periodic functions. In our exercise, the function \( f(x)=2\sin(x)+\sin(2x) \) involves both the sine and double angle sine functions.
Understanding these functions is essential because:

• They describe periodic oscillations, which are common in physics and engineering.
• Their derivatives and integrals have distinctive patterns, which are crucial for integrative calculations.
• Recognizing identities and properties—for example, \( \sin(2x) = 2\sin(x)\cos(x) \)—can simplify more complex expressions.

Mastery of trigonometric functions and their properties not only aids in integration tasks but also in understanding the behavior of the functions they create.