Problem 53 - Calculus of a Single Variable

Question

Integration as an Accumulation Process In Exercises \(51-54\) , find the accumulation function \(F\) . Then evaluate \(F\) at each value of the independent variable and graphically show the area given by each value of \(F .\) $$ F(\alpha)=\int_{-1}^{\alpha} \cos \frac{\pi \theta}{2} d \theta \quad(\text { a }) F(-1) \quad \text { (b) } F(0) \quad \text { (c) } F\left(\frac{1}{2}\right) $$

Step-by-Step Solution

Answer

The values of the function F at the three given points are F(-1) = -2/\( \pi \), F(0) = 0, and F(1/2) = 1. The graph of the function represents the varying areas under the curve of the cosine function

Step 1: Analyzing the given function

The given function is an integral of \( \cos \frac{\pi \theta}{2} \) from -1 to \( \alpha \). This function can be integrated using standard methods.

Step 2: Evaluating the integral

The antiderivative of \( \cos \frac{\pi \theta}{2} \) is \( F(\alpha) = 2\frac{sin(\frac{\pi \alpha}{2})}{\pi} \). According to the Fundamental Theorem of Calculus, when we want to evaluate an integral, we need to substitute the upper bound in the place of \( \theta \) in the antiderivative and subtract the result of substituting the lower bound.

Step 3: Evaluating F(-1)

Plugging \( \alpha = -1 \) in \( F(\alpha) = 2\frac{sin(\frac{\pi \alpha}{2})}{\pi} \), we get \( F(-1) = -2/\pi \).

Step 4: Evaluating F(0)

Similarly, setting \( \alpha = 0 \) in \( F(\alpha) = 2\frac{sin(\frac{\pi \alpha}{2})}{\pi} \), yields \( F(0) = 0 \).

Step 5: Evaluating F(1/2)

Substituting \( \alpha = 1/2 \) into \( F(\alpha) = 2\frac{sin(\frac{\pi \alpha}{2})}{\pi} \), you get \( F(1/2) = 1 \).

Step 6: Graphing

Use the 3 computed points (-1, -2/\pi), (0,0), and (1/2, 1) to graph the function \( F(\alpha) \). Note that the graph of the function corresponds to the areas under the curve of \( cos( \frac{\pi \theta}{2} )\) from -1 to \( \alpha \).

Key Concepts

Integration

Integration is one of the two main operations in calculus, the other being differentiation. It is essentially the process of finding the area under a curve, which can be immensely helpful in understanding how quantities accumulate.
In our exercise, we are asked to find the accumulation function \(F\) using integration. An accumulation function tells us how much a certain quantity (in this case, the area under the curve \( \cos \frac{\pi \theta}{2} \)) has accumulated as we move from one boundary, here \(\theta = -1\), to a variable boundary \(\alpha\).

• Integration is used to find this accumulated quantity for any point \(\alpha\).
• The integral here is set with a lower limit of \(-1\) and an upper limit of \(\alpha\).
• Use integration techniques to evaluate specific points, such as \(\alpha = 0\), to find corresponding values of the accumulation function.

Understanding the process of integration allows us to explore how quantities build up over a specified range, which is a pivotal concept in calculus and many applied fields.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus (FTC) is a key principle that links the processes of differentiation and integration. It states that differentiation and integration are inverse processes.
Let's break this down with our function \( F(\alpha) = \int_{-1}^{\alpha} \cos \frac{\pi \theta}{2} \, d\theta \). When we look at this integral, the FTC tells us that the process of integrating the function gives us all the possible accumulated areas \( F(\alpha) \).

• The first part of the FTC allows us to find an antiderivative of the function \( \cos \frac{\pi \theta}{2} \).
• The second part helps us evaluate the definite integral by substituting the boundary values into the antiderivative.

The FTC simplifies the process by telling us that to find the area (or accumulated function value), we need the difference between the antiderivative at the upper boundary \( \alpha \) and the lower boundary \(-1\). This powerful theorem bridges the gap between finding instantaneous rates of change and total accumulated change.

Accumulation Function

An accumulation function is the result of integrating a function over a range of values. It tells us how a function's integral builds up as we move across different values of \(\alpha\).
In our exercise, the function \( F(\alpha) = \int_{-1}^{\alpha} \cos \frac{\pi \theta}{2} \, d\theta \) signifies how accumulated area under the curve changes with \(\alpha\).

• As we substitute different values of \(\alpha\), such as \(\alpha = 0\) or \(\alpha = \frac{1}{2}\), we notice changes in \(F(\alpha)\).
• For \(\alpha = -1\), the accumulation starts as \(-2/\pi\).
• For \(\alpha = \frac{1}{2}\), the accumulation reaches \(1\).

This shows that an accumulation function doesn’t just tell us a sum but shows how the accumulation progresses as \(\alpha\) increases. It's a versatile tool giving insight into how total quantities build up in relation to another variable.

Definite Integral

A definite integral represents the accumulation of quantities, generally limited between two points — a lower and an upper bound. It is symbolized as \(\int_{a}^{b} f(x) \, dx\), where \(a\) and \(b\) are the limits of integration.
In our exercise, the definite integral from \(-1\) to \(\alpha\) for the function \(\cos \frac{\pi \theta}{2}\) provides us with precise accumulated values within these bounds.

• It is called 'definite' because it computes a specific numerical value (area, in geometrical terms) between set limits.
• By using these specific limits, the integral helps us understand the accumulation within a confined section of the domain.
• For instance, plugging different values into \(F(\alpha)\) yields exact integral results for those specific ranges.

Utilizing the concept of definite integrals aids us in calculating exact changes or quantities accumulated over a given interval, thus offering clarity and precision in analyzing functions.