Problem 3 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Put the following integrals in ascending order, placing \(<\) or \(=\) signs between them as appropriate. (Strategy: First determine which integrals are positive, which are negative, and which are zero.) (a) \(\int_{0}^{\pi} \sin (t) d t\) (b) \(\int_{-\pi}^{2 \pi} \sin (t) d t\) (c) \(\int_{-\pi}^{2 \pi} \cos (t) d t\) (d) \(\int_{0}^{\pi / 2} \cos (t) d t\) (e) \(\int_{0}^{\pi} \cos (2 t) d t\) (f) \(\int_{0}^{3 \pi / 2}|\sin (t)| d t\)

Step-by-Step Solution

Answer

In ascending order, the integrals are: (c) = (b) = (e) \(<\) (d) \(<\) (a) \(<\) (f).

Step 1: Evaluate each integral individually

(a) The integral \(\int_{0}^{\pi} \sin (t) dt\) represents the area under the curve of the sin function from 0 to \(\pi\). Since sine fluctuates between -1 and 1, this integral gives a positive result. Specifically, the integral equals 2. (b) The integral \(\int_{-\pi}^{2 \pi} \sin (t) dt\) takes the area under the sin function from -\(\pi\) to \(2 \pi\), which includes two full periods of the function. The property of the sine function is that the integral over a period is zero, so the result is zero. (c) The integral \(\int_{-\pi}^{2 \pi} \cos (t) dt\) behaves similarly to integral (b) because the cosine function also has a period of \(2\pi\). Thus, this integral is also zero. (d) The integral \(\int_{0}^{\pi / 2} \cos (t) dt\) represents the area under the cos function from 0 to \(\pi / 2\). Since cosine is at its maximum over this interval, the integral is positive and equals 1. (e) The integral \(\int_{0}^{\pi} \cos (2 t) dt\) involves a cosine function with a doubled argument, which reduces the period to \(\pi\). This means that over the interval from 0 to \(\pi\), it completes a full cycle, so the integral is zero. (f) The integral \(\int_{0}^{3 \pi / 2} |sin (t)| dt\) gives the area under the absolute value of the sine function from 0 to \(3\pi / 2\). Since the sine function is always non-negative when taken the absolute value, this integral gives a larger value than a plain sine integral would over the same interval. The result is 3.

Step 2: Comparing and Ordering

Now, compare the calculated values and order the functions in ascending order. We can safely ignore any non-unique values (equals 0 in this case); they are neither less than nor greater than the other functions. By reviewing the calculated values, this is the order: (c) = (b) = (e) \(<\) (d) \(<\) (a) \(<\) (f).

Key Concepts

Definite Integral

Understanding the definite integral is crucial for solving problems related to calculating the area under a curve, among other applications. In essence, a definite integral represents the total accumulation of a quantity, which we calculate over a certain interval. When given a function such as \( f(x) \), the definite integral from \( a \) to \( b \) is denoted by \( \int_{a}^{b} f(x) dx \). The process of finding this integral involves calculating the limit of Riemann sums as the number of intervals approaches infinity.

In the context of the exercise, we evaluated definite integrals to find the area under various trigonometric functions. For example, the integral of the sine function from \( 0 \) to \( \pi \) can be visualized as the area of the curve lying above the x-axis, giving us a positive result. Thus, understanding the geometric interpretation of the definite integral helps us compare the results obtained from different integrals, as was required in the exercise.

This concept becomes particularly important in physics and engineering, where it often represents quantities such as displacement or total work done.

Sine and Cosine Integrals

The sine and cosine functions are periodic and oscillate between their maximum and minimum values. An integral involving these functions over a symmetric interval that encompasses their complete cycles will result in zero. This is linked to the property that the positive and negative areas under these curves will cancel each out over one full period.

In terms of evaluating these integrals, there are specific identities that might simplify the process. For instance, an integral of cosine doubled (\(\cos(2t)\)) can be handled using the identity for cosine of a double angle. Moreover, understanding the behavior of sine and cosine functions within their respective periods was paramount to solving the exercise.

Let's consider a trigonometric integral from the exercise, \( \int_{0}^{\pi} \sin(t) dt \), which yields a positive value because it covers the area where the sine function is positive, from 0 to \( \pi \). Accurately understanding these concepts enables us to anticipate the outcome of trigonometric integrals and their applications, like predicting the behavior of alternating current in electrical engineering.

Absolute Value of Trigonometric Functions

The absolute value of a trigonometric function, like \( |\sin(t)| \), alters the function's typical behavior. Instead of oscillating above and below the x-axis, the absolute value ensures the function stays non-negative. Integrals of these modified functions, such as \( \int_{0}^{3\pi/2} |\sin(t)| dt \), no longer cancel themselves out over a period since all the areas are taken as positive.

The importance of this concept in applied mathematics and physics is undeniable. For instance, when computing work done against resistance, one might need to consider distances traveled without regard to direction, which is where the concept of taking the absolute value comes into play.

The exercise illustrates this beautifully, as it showed that the definite integral values change significantly when we consider the area under the curve of the absolute value of a trigonometric function. In this case, the result was larger than a standard integral, highlighting the impact absolute values have on the evaluation of integrals.