Problem 50 - Calculus

Question

Use the Substitution Rule for Definite Integrals to evaluate each definite integral. $$ \int_{0}^{1 / 2} \sin (2 \pi x) d x $$

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{1}{\pi} \).

Step 1: Identify the Substitution

To simplify the integration process, use a substitution to transform the integral. Let \( u = 2\pi x \). This substitution will change both the integrand and the limits of integration.

Step 2: Differentiate the Substitution

Calculate \( du \) in terms of \( dx \). Differentiate \( u = 2\pi x \) with respect to \( x \), giving \( \frac{du}{dx} = 2\pi \), which implies \( du = 2\pi \, dx \), or equivalently, \( dx = \frac{du}{2\pi} \).

Step 3: Change the Limits of Integration

Substitute the original limits of integration \( x = 0 \) and \( x = \frac{1}{2} \) to find the new limits for \( u \). When \( x = 0 \), \( u = 2\pi \times 0 = 0 \). When \( x = \frac{1}{2} \), \( u = 2\pi \times \frac{1}{2} = \pi \). Thus, the new limits for \( u \) are from 0 to \( \pi \).

Step 4: Substitute into the Integral

Substitute \( u \) and \( dx \) into the original integral. The integral becomes:\[\int_{0}^{\pi} \sin(u) \left( \frac{du}{2\pi} \right) = \frac{1}{2\pi} \int_{0}^{\pi} \sin(u) \, du\]

Step 5: Integrate with Respect to u

Evaluate the integral \( \int \sin(u) \, du \), which is \( -\cos(u) \). Therefore, the integral is given by:\[\frac{1}{2\pi} \left[-\cos(u)\right]_{0}^{\pi} = \frac{1}{2\pi} \left[-\cos(\pi) + \cos(0)\right]\]

Step 6: Evaluate the Definite Integral

Calculate \( -\cos(\pi) + \cos(0) \). Since \( \cos(\pi) = -1 \) and \( \cos(0) = 1 \), it follows:\[\frac{1}{2\pi} \left[1 + 1\right] = \frac{2}{2\pi} = \frac{1}{\pi}\]

Key Concepts

Definite Integrals

A definite integral, represented as \( \int_{a}^{b} f(x) \, dx \), gives the accumulated area under a curve from one point to another on the x-axis. In such an integral, both the integrand \( f(x) \) and the limits of integration \( a \) and \( b \) play crucial roles. The process involves evaluating the antiderivative of \( f(x) \) at these limits and then finding the difference.

Not only does it offer insights into total accumulated values, like distance or area, but it also allows us to determine quantities that gradually accumulate over a range. The definite integral turns a function's expression into a tangible quantity by summing up infinitesimal contributors between two specific values. Thus, while indefinite integrals offer a general form (complete with a constant of integration), definite integrals provide a precise number, representing a quantifiable outcome.

For example, in the exercise, once the substitution rule is applied and the new integral is evaluated from 0 to \( \pi \), it allows us to calculate the accumulated area under the curve \( \sin(2\pi x) \) between the original limits \( x = 0 \) and \( x = \frac{1}{2} \). The result of this particular integration is \( \frac{1}{\pi} \).

Trigonometric Integrals

Trigonometric integrals involve functions like sine, cosine, tangent, and others. Calculating these integrals usually requires knowledge of their derivatives and integrals, helping you simplify the process. For instance, the integral \( \int \sin(u) \, du \) is a basic trigonometric integral, known to be \( -\cos(u) \).

The original problem includes \( \sin(2\pi x) \), a trigonometric expression. Transforming the variable using substitution (\( u = 2\pi x \), in this case) simplifies the integral to a more straightforward form. Recognizing these integral forms and knowing their antiderivatives are essential skills for solving calculus problems effectively. It becomes especially important for tackling problems that might feature complex trigonometric relationships or where integration bounds complicate straightforward calculations.

Trigonometric integrals often come up in physics and engineering where periodic functions model wave forms, oscillations, and other repeating processes. Understanding how to solve these integrals can significantly aid in handling real-world scenarios, making the abstract mathematics tangible and applicable.

Limits of Integration

The limits of integration define the part of the x-axis over which you perform your definite integral. They are the values at which you start and stop accumulating the function's area under the curve. In mathematical notation, the integral limits are expressed as the bottom and top numbers of the integral symbol, like \( \int_{a}^{b} \). Changing these limits through substitution is an important technique when solving integrals.

In our example, the substitution \( u = 2\pi x \) requires adjusting the original limits of \( x = 0 \) and \( x = \frac{1}{2} \). To find the new limits for \( u \), we substitute these x-values into the substitution formula: \( u = 2\pi \times 0 = 0 \) and \( u = 2\pi \times \frac{1}{2} = \pi \).

• Original lower limit: \( x = 0 \) changes to \( u = 0 \).
• Original upper limit: \( x = \frac{1}{2} \) changes to \( u = \pi \).

By changing these limits, the mathematics seamlessly shift to a new variable, often making the integral simpler to solve.

Proper alignment of limits with substitution is crucial. Failing to adjust them properly for the new variable can lead to erroneous results. Thus, it's essential to carefully transform these bounds alongside the integrand for accurate calculations and interpretations.