Problem 41 - Essential Calculus

Question

In Exercises 41-44, (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 calculated by integrating the function \(f(x) = 2 \sin x + \sin 2x\) between \(x = 0\) and \(x = \pi\) and confirmed using a graphical calculator is the correct area under the curve.

Step 1 - Sketching the Region

Use a graphing utility to sketch the function \(f(x) = 2 \sin x + \sin 2x\) between 0 and \(\pi\). The sketch must also include the line \(y = 0\) (the x-axis), as this is the lower bound of the region of interest.

Step 2 - Compute Area

The area under the curve from \(x = 0\) to \(x = \pi\) is given by the definite integral \(\int_{0}^{\pi} f(x) \, dx\). In order to find the area, calculate the definite integral \(\int_{0}^{\pi} [2 \sin x + \sin (2x)]dx\). Remember that the integral of \(\sin x\) is \(-\cos x\) and the integral of \(\sin (2x)\) is \(-\frac{1}{2}\cos(2x)\). Substitute the upper and lower limits of integration to find the total area.

Step 3 - Verify Result

Finally, check the result from Step 2 using the integral feature of the graphing utility. The definite integral \(\int_{0}^{\pi} [2 \sin x + \sin (2x)]dx\) must be evaluated using the utility. If this result matches the previous computation, then the answer can be trusted, otherwise, the calculations need to be reviewed.

Key Concepts

Graphing Utility

When tackling calculus problems, a graphing utility is an incredible asset for visual learners. It turns complex equations into understandable visuals, allowing students to see the shape and behavior of functions like the sine wave. This visual representation is especially helpful when identifying the region bounded by function curves and axis lines.

When used correctly, a graphing utility can display the function's behavior between any two points, such as between 0 and \(\pi\) for the exercise provided. It assists in picturing the area that needs to be calculated, making the otherwise abstract concept of integration more tangible. It's also a useful tool for confirming solutions, as the area calculation can be quite complex and prone to manual errors.

Area Under Curve

The area under the curve is a fundamental concept in calculus representing the accumulation of quantities. When the curve represents a rate of change, the area under it from one point to another represents the total change over that interval. Calculating the area under the sine curve from 0 to \(\pi\), as in our exercise, is particularly useful in physics and engineering, where sine functions often model periodic phenomena.

By learning to calculate the area under curves, students gain insight into real-world applications of calculus, such as determining distances traveled or the average value of a function over an interval.

Integration

Integration is a cornerstone of calculus, used to find areas, volumes, central points, and many other useful things. But more than just these applications, integration represents the reverse process of differentiation. Where differentiation breaks things into small pieces to find how they change, integration combines small pieces to find how much there is.

The definite integral, with its upper and lower limits, specifies the interval over which we want to 'sum up' the function. For our exercise, the definite integral \(\int_{0}^{\pi} f(x) \, dx\) gives us the precise area under \(f(x)\) from 0 to \(\pi\), making it a fundamental concept for students to master.

Sine Function Integration

The sine function, denoted as \(\sin x\), is prevalent in various scientific fields due to its periodic nature. Integrating a sine function, which occurs in our example as \(2 \sin x + \sin 2x\), involves understanding its antiderivative. The integral of \(\sin x\) is \(\-\cos x\), stemming from the fact that the derivative of \(\cos x\) is \(\-\sin x\). For functions like \(\sin(2x)\), a coefficient in the argument introduces a necessary adjustment, resulting in an integral of \(\-\frac{1}{2}\cos(2x)\).

Grasping this concept is crucial for students as it extends beyond simple sine functions to more complex trigonometric integrals. Understanding how to integrate these functions paves the way for solving a broad range of problems involving waves, oscillations, and many natural phenomena governed by trigonometric relations.

Calculus Education

Calculus education does more than teach students how to manipulate mathematical symbols. It offers a new way of thinking about change, motion, and growth. By breaking down complex concepts like integration and the area under a curve into bite-sized, easy-to-understand pieces, students can build a strong foundation for advanced mathematical, scientific, and engineering tasks.

Moreover, the step-by-step methodology, as shown in the exercise provided, helps students to follow a logical progression from problem recognition, through visualization with a graphing utility, to solution verification. Encouraging comprehension through visualization and practical application ensures that students not only know how to get the answer but also understand the underlying principles and their real-world implications.