Problem 50 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 35-62, 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 value of the integral is \( \frac{1}{\pi} \).

Step 1: Identify the Substitution

We start by choosing a substitution for the integral. Looking at \( \int \sin(2\pi x) \, dx \), the expression inside the sine function, \( 2\pi x \), suggests the substitution \( u = 2\pi x \).

Step 2: Compute the Differential

Compute the differential \( du \) by differentiating \( u = 2\pi x \). This gives \( du = 2 \pi \, dx \), or rearranging, \( dx = \frac{du}{2\pi} \).

Step 3: Change the Limits of Integration

Substitute the limits of integration. When \( x = 0 \), \( u = 2\pi \times 0 = 0 \). When \( x = \frac{1}{2} \), \( u = 2\pi \times \frac{1}{2} = \pi \). Thus, the limits change from \( x = 0 \) to \( x = \frac{1}{2} \) to \( u = 0 \) to \( u = \pi \).

Step 4: Substitute in the Integral

Substitute \( u \) and \( dx \) in the integral: \[ \int_{0}^{1/2} \sin(2\pi x) \, dx = \int_{0}^{\pi} \sin(u) \cdot \frac{du}{2\pi} = \frac{1}{2\pi} \int_{0}^{\pi} \sin(u) \, du. \]

Step 5: Evaluate the Integral

Find the antiderivative of \( \sin(u) \), which is \( -\cos(u) \). Evaluate this from \( 0 \) to \( \pi \): \[ \frac{1}{2\pi} \left[ -\cos(u) \right]_{0}^{\pi} = \frac{1}{2\pi} [-\cos(\pi) + \cos(0)] = \frac{1}{2\pi} [-(-1) + 1] = \frac{1}{2\pi} [2] = \frac{1}{\pi}. \]

Step 6: Conclusion

The value of the definite integral \( \int_{0}^{1 / 2} \sin (2 \pi x) \, dx \) is \( \frac{1}{\pi} \).

Key Concepts

Substitution Rule

The Substitution Rule is a technique used to simplify finding the value of definite integrals, especially when dealing with complex expressions under the integral sign. To effectively apply this rule, we choose a substitution that simplifies the original integrand.

• Identify a substitution that turns the integrand into a simpler form, such as choosing a variable for a composite expression.
• Calculate the differential of the new variable, ensuring to express the differential of the original variable in terms of the new one.
• Change the integration limits to reflect your substitution, transforming the problem into a new interval.
• Replace the original integrand and limits with your substitutions in the integral, simplifying to a form that is easier to evaluate.

For example, in the integral \( \int \sin(2\pi x) \, dx \), you choose \( u = 2\pi x \), simplifying the integration process. Remember that your new integral must respect the terms of differentials and limits transformations, setting the groundwork for finding the antiderivative more straightforwardly.

Antiderivatives

Antiderivatives, also known as indefinite integrals, refer to functions whose derivative gives back the original function within the integrand. This is the reverse process of differentiation.

• The primary goal when dealing with antiderivatives in definite integrals is to first determine the indefinite integral before applying the established limits of integration.
• For trigonometric functions like \( \sin(u) \), the antiderivative in terms of \( u \) would be \( -\cos(u) \), representing the family of functions whose derivative equals \( \sin(u) \).

Once you have the antiderivative, you plug in the upper and lower limits of the transformation into this function, returning the difference to obtain the definite integral's result. In the provided example, the integral of \( \sin(u) \) across transformed limits leads to a computing process resulting in an expression evaluated from \( u = 0 \) to \( u = \pi \), ultimately giving a numeric value. This reflects a broader principle in integral calculus of translating areas under curves into exact numbers.

Trigonometric Functions

Trigonometric functions like sine and cosine play a central role in integration due to their continuous and periodic nature. They are widely encountered in various scientific fields, requiring precise understanding during calculations.

• These functions help model and solve real-world problems involving oscillation, such as waves and circular motion.
• In calculus, recognizing integral forms of trigonometric functions enables methods like substitution and integration by parts, providing templates to approach calculations systematically.

For instance, the integral \( \int \sin(2\pi x) \, dx \) involves identifying a compound angle within the sine function to simplify the integral through substitution. Understanding these functions' properties, like their periodicity and symmetry, greatly aids in calculating antiderivatives and definite integrals. Moreover, interpreting results with respect to these functions considers their positive and negative behavior over specific intervals, ensuring that your calculations are accurate. Their applications stretch across engineering, physics, and everyday mathematics, underscoring the significance in mastering these concepts.