Problem 51 - Essential Calculus

Question

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

\(F(-1) = 0\), \(F(0) = \frac{2}{\pi}\) and \(F(\frac{1}{2}) = \frac{2}{\pi}(1+ \sqrt{2}/2)\)

Step 1: Compute the accumulation function F(α)

Based on the fundamental theorem of calculus, we compute the integral from -1 to α of \(\cos(\frac{π\theta}{2})\) dΘ. When we do this, we get: \(F(α) = \frac{2}{π}[\sin(\frac{πα}{2}) + \sin(\frac{π}{2})] \)

Step 2: Evaluate F(α) at α=-1

Substitute α = -1 into \(F(α)\). So we have \(F(-1) = \frac{2}{π}[\sin(-\frac{π}{2}) + \sin(\frac{π}{2})] = 0\)

Step 3: Evaluate F(α) at α=0

Substitute α = 0 into \(F(α)\). This gives: \(F(0) = \frac{2}{π}[\sin(0) + \sin(\frac{π}{2})] = \frac{2}{π}\)

Step 4: Evaluate F(α) at α=1/2

We then substitute α = 1/2 into \(F(α)\). This gives: \(F(\frac{1}{2}) = \frac{2}{π}[\sin(\frac{π}{2}) + \sin(\frac{π}{4})] = \frac{2}{π}(1+ \sqrt{2}/2)\)

Step 5: Graphical Representation

Plot these points (-1, 0), (0, 2/π), (1/2, 2/π * (1+sqrt(2)/2)) on a graph. Join them using smooth curves. The graph is the graphical representation of F(α)

Key Concepts

Accumulation Function

An accumulation function, like the one presented here for integration purposes, is a mathematical tool used to measure the total accumulated change of a function over an interval. In simpler terms, it tells us how much 'accumulation' has occurred from one point to another. This is particularly useful in calculus when we're interested in understanding the overall effect described by a function simplified through an integral.

For this exercise, we have the function represented as:

• \( F(\alpha) = \int_{-1}^{\alpha} \cos \left(\frac{\pi \theta}{2}\right) d \theta \)

To find the accumulation function \( F(\alpha) \), you use the definite integral over the range from -1 to \( \alpha \). Essentially, you are summing up the areas under the curve of \( \cos \left( \frac{\pi \theta}{2} \right) \) within this interval. The fundamental theorem of calculus allows us to transition smoothly from integration to evaluation—a key aspect in calculus questions like this one.

Definite Integral

The definite integral is a core part of this question and, importantly, it is what transforms the function \( \cos \left( \frac{\pi \theta}{2} \right) \) into a tangible value, \( F(\alpha) \), over the specified interval. In calculus, definite integrals are used to compute the integral of a function between two bounds, providing the net area under the curve between these points.

Using the definite integral, we can compute the exact values of the accumulation function. For example, the solution computes this integral at specific values of \( \alpha \): -1, 0, and \( \frac{1}{2} \). Here’s a quick rundown:

• At \( \alpha = -1 \): the area is zero. This is expected because you are evaluating from the lower bound of the interval, which results in no accumulation.
• At \( \alpha = 0 \): the accumulation yields \( \frac{2}{\pi} \). This shows a specific amount of area collected under the curve from \(-1\) to \( 0 \).
• At \( \alpha = \frac{1}{2} \): we find \( F(\frac{1}{2}) = \frac{2}{\pi}(1+ \sqrt{2}/2) \), indicating increased accumulation as the interval extends further.

This calculation process reflects the essence of the definite integral in summarizing the "whole" captured area.

Graphical Representation

Visualizing functions and their integrals plays a crucial role in understanding their behavior. Graphical representation helps translate mathematical results into more intuitive insights. In this exercise, plotting the evaluated points of the accumulation function \( F \) provides a visual reference for how the accumulation changes as \( \alpha \) changes.

The function values at \( \alpha = -1 \), \( \alpha = 0 \), and \( \alpha = \frac{1}{2} \) give us coordinates that can be plotted as

• (-1, 0)
• (0, \( \frac{2}{\pi} \))
• (\( \frac{1}{2} \), \( \frac{2}{\pi} (1+\sqrt{2}/2) \))

These points are connected smoothly to show the path the function takes as \( \alpha \) changes. This graphical plot gives concrete meaning to these numbers and helps us see the behavior and growth pattern of the accumulation function over the interval. It’s a clear representation of the theoretical calculations in a format that is easier to digest visually.