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Integration as an Accumulation Process Exercises \(53 - 56\) , 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 the independent variable. $$F ( \alpha ) = \int _ { - 1 } ^ { \alpha } \cos \frac { \pi \theta } { 2 } d \theta \quad$$ (a) \(F ( - 1 ) \quad\) (b) \(F ( 0 ) \quad\) (c) \(F \left( \frac { 1 } { 2 } \right)\)

Short Answer

Expert verified
The accumulation function \(F(\alpha) = \frac{2}{\pi}\left[\sin\left(\frac{\pi \alpha}{2} \right) + 1\right]\). The values of this function for the given points are (a) 0, (b) \(\frac{2}{\pi}\), and (c) \(\frac{2}{\pi}(\frac{\sqrt{2}}{2}+1)\)

Step by step solution

01

Identify the integral in question and find its antiderivative

The given integral is \(\int \cos \left( \frac{\pi \theta}{2} \right) d \theta\). We proceed to find its antiderivative. The antiderivative of \(\cos u\) is \( \sin u \). Using the substitution technique for integrals, we let \(u = \frac{\pi \theta}{2}\), compute \(du= \frac{\pi}{2}d\theta\). Rearranging for \(d \theta\), we get \(d\theta = \frac{2 du}{\pi}\). We substitute these into the integral and compute the antiderivative as \(\frac{2}{\pi} \sin \left(\frac{\pi \theta}{2} \right)\).
02

Apply the limits of integration

We now evaluate our antiderivative at the limits \(\alpha\) and -1. This results in \(F(\alpha) = \frac{2}{\pi}\left[\sin\left(\frac{\pi \alpha}{2} \right) - \sin\left(\frac{\pi(-1)}{2}\right)\right]\). Since \(\sin \left( \frac{\pi(-1)}{2} \right)= \sin\left(-\frac{\pi }{2}\right)\) equals -1, our accumulation function simplifies to \( F(\alpha) = \frac{2}{\pi}\left[\sin\left(\frac{\pi \alpha}{2} \right) + 1\right]\)
03

Evaluate the function at specific points

Now we will find the values of \(F\) for the specific points (a) -1, (b) 0, and (c) \(\frac{1}{2}\). \n(a) Substituting \(\alpha = -1\), we get \(F(-1) = \frac{2}{\pi}\left[\sin\left(\frac{-\pi }{2} \right) + 1\right] = 0\), \n(b) For \(\alpha = 0\), \(F(0) = \frac{2}{\pi}\left[\sin\left(0 \right) + 1\right] = \frac{2}{\pi}\), \n(c) And for \(\alpha = \frac{1}{2}\), \(F\left(\frac{1}{2} \right) = \frac{2}{\pi}\left[\sin\left(\frac{\pi }{4} \right) + 1\right] = \frac{2}{\pi}(1+ \sqrt{2}/2)\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Definite Integral
A definite integral is a powerful mathematical tool that allows us to compute the total accumulation of quantities described by a function, between two specified values called limits of integration. Consider it like finding the total area under a curve, bounded by two points on the x-axis.

For example, the exercise involves evaluating a definite integral from -1 to \( \alpha \). The role of this integral is to track the total change in the function \( \cos \frac{\pi \theta}{2} \) over that interval. To calculate the definite integral, we first find the antiderivative (a "reverse" of the derivative) of the function and then apply the limits of integration to determine the accumulation function \( F(\alpha) \). This process involves mathematical techniques that form the core of calculus.

In practical terms:
  • The lower limit (in the exercise, \( -1 \)) is where computation of the area starts.
  • The variable \( \alpha \) marks the upper limit of the integral, where the calculation stops.
  • Evaluating the definite integral gives tangible numerical results for the area, hence providing insight into the function's behavior between two points.
Antiderivative
An antiderivative is a crucial concept in calculus, often considered the inverse operation of differentiation. While differentiation breaks down a function into its rates of change, finding the antiderivative or "integral" stitches these changes back together to approximate the original function.

To solve the exercise, we compute the antiderivative of \( \cos \frac{\pi \theta}{2} \). The antiderivative of a trigonometric function like \( \cos u \) is \( \sin u \), since derivative of \( \sin u \) with respect to \( u \) is \( \cos u \). This relationship is integral to understanding how changes are accumulated.

In practical execution:
  • Recognize trigonometric patterns to simplify the integration process.
  • After substitution (which we’ll discuss shortly), transform the integral into standard formats.
  • The computed antiderivative forms the basis by which the limits of integration can be applied.
Understanding and finding antiderivatives is essential for the integral to correctly reflect accumulated quantities.
Integration by Substitution
Integration by substitution is a handy method for simplifying complex integrals. By changing variables, we streamline integration processes, making them much more manageable akin to the chain rule in differentiation.

In the exercise, we set \( u = \frac{\pi \theta}{2} \) to simplify the integral of \( \cos \frac{\pi \theta}{2} d\theta \). By this logic for substitution:
  • Differentiate \( u \) to find \( du \), ensuring the original variable \( \theta \) is completely replaced in terms of \( u \).
  • Solve for \( d\theta \) to substitute back and simplify the integral in terms of \( du \).
  • This technique transforms a difficult integral into an easier form, making computations straightforward.
Integration by substitution is reminiscent of algebraic manipulation, offering a resourceful tool for unraveling and solving trigonometric and polynomial integrals efficiently.
Trigonometric Integrals
Trigonometric integrals involve functions like sine, cosine, and tangent, which occur frequently in calculus. Due to their periodic nature, calculating these integrals requires familiarity with specific identities and their properties.

For the given function \( \cos \frac{\pi \theta}{2} \), using known antiderivatives is key. Basic identities such as;
  • The antiderivative of \( \cos u \) is \( \sin u \).
  • Understanding the periodicity and symmetry of trigonometric functions can simplify evaluations.
These integrals leverage trigonometric identities and substitution to effectively calculate area under curves defined by sinusoidal functions. Mastery of trigonometric fundamentals equips you to tackle more complex calculus problems, as seen in many real-world phenomena, from wave patterns to oscillations.

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Most popular questions from this chapter

Modeling Data A pond is approximately circular, with a diameter of 400 feet. Starting at the center, the depth of the water is measured every 25 feet and recorded in the table (see figure). $$\begin{array}{|c|c|c|c|}\hline x & {0} & {25} & {50} \\ \hline \text { Depth } & {20} & {19} & {19} \\ \hline\end{array}$$ $$\begin{array}{|c|c|c|c|}\hline x & {75} & {100} & {125} \\ \hline D e p t h & {17} & {15} & {14} \\ \hline\end{array}$$ $$\begin{array}{|c|c|c|c|}\hline x & {150} & {175} & {200} \\ \hline \text { Depth } & {10} & {6} & {0} \\ \hline\end{array}$$ $$\begin{array}{l}{\text { (a) Use the regression capabilities of a graphing utility to find }} \\ {\text { a quadratic model for the depths recorded in the table. Use }} \\ {\text { the graphing utility to plot the depths and graph the model. }}\end{array}$$ $$\begin{array}{l}{\text { (b) Use the integration capabilities of a graphing utility and }} \\ {\text { the model in part (a) to approximate the volume of water }} \\ {\text { in the pond. }}\end{array}$$ $$\begin{array}{l}{\text { (c) Use the result of part (b) to approximate the number of }} \\ {\text { gallons of water in the pond. (Hint: } 1 \text { cubic foot of water }} \\ {\text { is approximately } 7.48 \text { gallons.) }}\end{array}$$

Graphical and Numerical Reasoning Consider the region bounded by the graphs of \(y=x^{2 n}\) and \(y=b,\) where \(b>0\) and \(n\) is a positive integer. (a) Sketch a graph of the region. (b) Set up the integral for finding \(M_{y}\) . Because of the form of the integrand, the value of the integral can be obtained without integrating. What is the form of the integrand? What is the value of the integral and what is the value of \(\overline{x} ?\) (c) Use the graph in part (a) to determine whether \(\overline{y}>\frac{b}{2}\) or \(\overline{y}<\frac{b}{2} .\) Explain. (d) Use integration to find \(\overline{y}\) as a function of \(n .\) (e) Use the result of part (d) to complete the table. (f) Find \(\lim _{n \rightarrow \infty} \overline{y}\) (g) Give a geometric explanation of the result in part (f).

Modeling Data A draftsman is asked to determine the amount of material required to produce a machine part (see figure). The diameters \(d\) of the part at equally spaced points \(x\) are listed in the table. The measurements are listed in centimeters. $$\begin{array}{|c|c|c|c|c|c|c|}\hline x & {0} & {1} & {2} & {3} & {4} & {5} \\\ \hline d & {4.2} & {3.8} & {4.2} & {4.7} & {5.2} & {5.7} \\\ \hline\end{array}$$$$\begin{array}{|c|c|c|c|c|c|}\hline x & {6} & {7} & {8} & {9} & {10} \\ \hline d & {5.8} & {5.4} & {4.9} & {4.4} & {4.6} \\\ \hline\end{array}$$ (a) Use the regression capabilities of a graphing utility to find a fourth-degree polynomial through the points representing the radius of the machine part. Plot the data and graph the model. (b) Use the integration capabilities of a graphing utility to approximate the volume of the machine part.

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