Problem 7 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Put the following four integrals in ascending order (from smallest to largest). Explain, using graphs, how you can be sure that the order you gave is correct. $$ \int_{0}^{\pi} \sin t d t, \quad \int_{0}^{\pi} 2 \sin t d t, \quad \int_{0}^{\pi} \sin (2 t) d t, \quad \int_{\pi}^{0} \sin t d t $$

Step-by-Step Solution

Answer

Arranging the integrals in ascending order gives: \(\int_{\pi}^{0} \sin t dt, \int_{0}^{\pi} \sin (2 t) dt, \int_{0}^{\pi} \sin t dt, \int_{0}^{\pi} 2 \sin t dt\)

Step 1: Integration of \(\int_{0}^{\pi} \sin t d t\)

The definite integral of \(\sin t\) from 0 to \(\pi\) can be calculated using the antiderivative of \(\sin t\) which is \(-\cos t\). Thus, the definite integral becomes \(-\cos t \Big|_0^\pi = -\cos(\pi) - -\cos(0) = -(-1) - -1 = 2\)

Step 2: Integration of \(\int_{0}^{\pi} 2 \sin t dt\)

Here, the function to integrate is \(2sin(t)\). Antiderivative gives us \(-2\cos t\), giving us a result \(-2\cos t \Big|_0^\pi = -2(-1) - -2(1) = 4\)

Step 3: Integration of \(\int_{0}^{\pi} \sin (2 t) d t\)

The definite integral can be calculated with the substitution rule by using \(u=2t\). After integrating, we would get \(-\frac{1}{2}\cos (2t)\Big|_0^\pi = -\frac{1}{2}(0 - 1) = 0.5\)

Step 4: Integration of \(\int_{\pi}^{0} \sin t d t\)

This integral is similar to the first one, but the limits of integration are reversed. It is the negative of the first integral and equals to \(-2\)

Step 5: Ordering the integrals

Now arrange these calculations in ascending order: \(\int_{\pi}^{0} \sin t d t < \int_{0}^{\pi} \sin (2 t) d t < \int_{0}^{\pi} \sin t d t < \int_{0}^{\pi} 2 \sin t dt\)

Key Concepts

Integration Techniques

Integration, a fundamental component of calculus, is a method of finding the area under a curve or between curves. Various integration techniques exist to solve integrals, which are essential in calculating definite integrals like those given in the exercise. It is often not enough to know the rules; understanding when and how to apply them is crucial.

One common technique is the antiderivative method, where you find a function that, when differentiated, gives you the integrand. This method was used for the simplest integral in the exercise, \( \int_{0}^{\pi} \sin t dt \). An understanding of basic trigonometric integrals is paramount here, as it allows for quick computation of this integral and others involving basic trigonometric functions.

Other techniques include substitution, which was applied in \( \int_{0}^{\pi} \sin (2 t) dt \) by substituting \( u = 2t \). There are also methods like integration by parts, partial fractions, and trigonometric substitution, each of which might be the most efficient approach depending on the integrand's form. Recognizing the correct technique often comes with practice and familiarity with different types of functions and their common antiderivatives.

Antiderivative

The term antiderivative refers to the inverse operation of taking a derivative. In other words, finding an antiderivative means identifying a function whose derivative is the given function. This concept is heavily used in solving definite integrals, as shown in the step-by-step solution. For example, the antiderivative of \( \sin t \) is \( -\cos t \), and this relationship is utilized to evaluate the definite integral from \( 0 \) to \( \pi \).

Understanding the concept of an antiderivative is essential because it lays the foundation for evaluating the area under a curve - the overarching goal of definite integration. Notably, the antiderivative is not unique; it can differ by a constant, which is why when evaluating a definite integral, the difference in the values of the antiderivative at the boundary points is taken.

The calculation of the antiderivative is also where students often improve their exercise by applying constants of integration appropriately, simplifying expressions before integrating, and taking advantage of symmetry in functions. For instance, integrating \( 2 \sin t \) involves pulling out constants to simplify integration, a valuable tactic in solving such problems.

Trigonometric Integrals

Integrals that involve trigonometric functions, such as \( \sin x \) or \( \cos x \), are known as trigonometric integrals. There's a suite of strategies to tackle them, some of which are showcased in the exercise's solutions. Trigonometric integrals often rely on identities or transformations to simplify the functions for integration.

For example, \( \int_{0}^{\pi} \sin t dt \) is straightforward due to the basic sine function. In contrast, \( \int_{0}^{\pi} \sin (2 t) dt \) involves a frequency modification represented by the \( 2t \) inside the sine function. This adjustment necessitates a substitution method for simplification. Moreover, recognizing the periodic properties of trigonometric functions can be particularly helpful, as integrals over a full period of a sine or cosine function will always equal zero, which is reflected in the exercise when the integral of \( \sin(2t) \) computed to \( 0.5 \) due to the bounds coinciding with the sine function's period.

In practice, these types of integrals are vital in physics and engineering, where they appear in wave functions and harmonic motion analyses. Familiarity with their properties and various techniques for solving them is therefore incredibly useful for students.