Question

Finding Area Show that if \(k\) is a positive constant, then the area between the \(x\) -axis and one arch of the curve \(y=\sin k x\) is always $$2 / k . \quad \int_{0}^{\pi / 2} \sin k x d x=\frac{2}{k}$$

Short answer

The area between the x-axis and one arch of the curve \( y = \sin kx \) is \( \frac{2}{k} \)

Step-by-step solution

Forming the Integral

To represent the area under one arch of the curve \( y = \sin kx \) from 0 to \( \frac{\pi}{2k} \), the definite integral to be calculated is: \( \int_{0}^{\pi / 2k} \sin kx \, dx \)

Evaluating the Integral

Next, evaluate this integral. The antiderivative of \( \sin kx \) is \( -\frac{1}{k} \cos kx \). Apply the Fundamental Theorem of Calculus to get the definite integral: \( \left[ -\frac{1}{k} \cos kx \right]_{0}^{\pi / 2k} \) which simplifies to \( -\frac{1}{k} \cos \left( \frac{\pi}{2} \right) - \left( -\frac{1}{k} \cos 0 \right) = -0 - \left( -\frac{1}{k} \right) = \frac{1}{k} \)

Finding Double Area Under the Curve

Above result is for half period of the sine function, starting from 0 to \( \frac{\pi}{2k} \). To get the area between the x-axis and one entire arch of the curve \( y = \sin kx \), i.e. from 0 to \( \frac{\pi}{k} \), the area will be twice of above, hence double the result to get: \( 2*\frac{1}{k} = \frac{2}{k} \)

Key concepts

Definite Integral

The definite integral is a fundamental concept in calculus, representing the accumulation of quantities and often the area under a curve. Suppose you're considering the curve represented by the function \( f(x) \), and you want to find the area under this curve between two points, \( a \) and \( b \) on the x-axis. This is given by the definite integral, which can be notated as \( \int_{a}^{b} f(x) \, dx \).
When calculating the area under the curve of a sine function, like \( y = \sin(kx) \), between \( x = 0 \) and \( x = \frac{\pi}{2k} \), we use this notion of definite integral. For positive constants \( k \), the definite integral helps us determine the precise area for a segment of the sine wave, ensuring a clear and easy understanding of spatial relationships within trigonometric functions.

• The lower limit of integration, \( a \), is the starting point on the x-axis.
• The upper limit of integration, \( b \), is the ending point on the x-axis.
• The function \( f(x) \) represents the height of the curve at any point \( x \).
• The integral sign \( \int \) symbolizes the aggregation of infinitely small areas from \( a \) to \( b \).

By understanding the definite integral, students can grasp how to compute areas for more complex curves beyond the basic shapes they're familiar with.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus (FTC) bridges the two central operations of calculus: differentiation and integration. It states that if a function \( f \) is continuous on the interval \( [a, b] \), and \( F \) is an antiderivative of \( f \) on this interval, then:
\( \int_{a}^{b} f(x) \, dx = F(b) - F(a) \).
The FTC makes it possible to calculate definite integrals without the need to sum an infinite series of rectangles under a curve - a process that would be both impractical and impossible to do with perfect accuracy. Instead, it allows us to evaluate the antiderivative of the function at the upper and lower limits of integration and simply subtract the former from the latter.

• FTC Part 1: Connects the concept of the derivative with the antiderivative.
• FTC Part 2: Enables the computation of definite integrals using antiderivatives.

In practice, when using the FTC for our textbook example, we evaluate the antiderivative of \( \sin(kx) \) at the bounds \( \frac{\pi}{2k} \) and \( 0 \) to find the area under one arch of the sine curve between these points. This theorem vastly simplifies the calculation of areas under curves, which is foundational for applications in physics, engineering, and beyond.

Sine Function Integration

Sine function integration is a specific instance where the function to be integrated is the sine function. In our example, we integrate the function \( y = \sin(kx) \) over a segment of its period. The integral of \( \sin(kx) \) over a certain interval can tell us the total area between the function's curve and the x-axis over that interval.

Why Know Sine Function Integration?

Integrating sine functions is important in physics for computing quantities like work and energy, where oscillatory functions frequently occur; in engineering for signal processing; and even in economics for cyclical patterns analysis. As the sine function is periodic, integrating it over an entire period results in an area of zero, since the positive and negative parts cancel each other out. However, integrating over a half-period, as in our exercise, gives the area under one 'hump' or 'trough' of the sine wave.
The computational process involves finding the antiderivative of the sine function, which, for \( \sin(kx) \), is \( -\frac{1}{k}\cos(kx) \). By applying the FTC, we can easily determine the definite integral of the function within the given limits.

Antiderivatives

An antiderivative of a function \( f \) is a function \( F \) whose derivative is \( f \). In other words, if \( F'(x) = f(x) \), then \( F \) is an antiderivative of \( f \). There can be infinitely many antiderivatives for any given function, and they are represented as \( F(x) + C \), where \( C \) is the constant of integration. Determining the antiderivative is the process of integration or finding the 'inverse' of differentiation.
In our exercise, the task is to find the antiderivative of \( y = \sin(kx) \), which is the function whose derivative is \( \sin(kx) \). This antiderivative is utilized to compute the definite integral of the sine function over a specified interval. It's critical to recognize that antiderivatives are the key tool allowing us to solve definite integrals without summing infinitesimal quantities directly, making them instrumental in solving a wide range of problems in mathematics and applied sciences.

• Understanding antiderivatives is essential for solving integration problems.
• Every continuous function has an antiderivative, which can be used to find areas under curves.

Antiderivatives provide a gateway to reversing the process of differentiation, and consequently, they are foundational to much of calculus, especially in understanding the behavior of functions and the areas they bound.