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Chapter 5: Problem 86

a. Show that if \(f\) is odd on \([-a, a],\) then $$\int_{-a}^{a} f(x) d x=0$$ b. Test the result in part (a) with \(f(x)=\sin x\) and \(a=\pi / 2\)

Short Answer

Expert verified
The integral of an odd function over equal bounds is zero; verified with \(\sin x\) on \([-\pi/2, \pi/2]\).

Step by step solution

01

Understand the Definition of an Odd Function

A function \(f\) is called odd if for every \(x\) in its domain, \(f(-x) = -f(x)\). This property will be essential for evaluating the integral.
02

Evaluate the Symmetric Property of the Integral

For an odd function \(f\) over the interval \([-a, a]\), the integral can be split as follows:\[ \int_{-a}^{a} f(x) \, dx = \int_{-a}^{0} f(x) \, dx + \int_{0}^{a} f(x) \, dx. \]
03

Use the Odd Function Property

Change variables in the first integral by setting \(u = -x\), then \(du = -dx\). Change limits from \(-a\) to \(0\) to \(a\) and \(0\):\[ \int_{-a}^{0} f(x) \, dx = \int_{a}^{0} f(-u) (-du) = \int_{0}^{a} f(-u) \, du. \]
04

Simplify Using Odd Function

Since \(f(-u) = -f(u)\) (because \(f\) is odd), this becomes:\[ \int_{0}^{a} f(-u) \, du = -\int_{0}^{a} f(u) \, du. \]
05

Combine Integrals

Substitute back into the expression:\[ \int_{-a}^{a} f(x) \, dx = -\int_{0}^{a} f(u) \, du + \int_{0}^{a} f(x) \, dx = 0. \]
06

Test the Result with \(f(x) = \sin x\) and \(a = \pi/2\)

The function \(f(x) = \sin x\) is odd since \(\sin(-x) = -\sin(x)\). Calculate:\[ \int_{-\pi/2}^{\pi/2} \sin x \, dx = 0 \]due to the proven property in Step 5.

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Key Concepts

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

Definite Integrals
Definite integrals are a core concept in calculus used to compute the accumulation of quantities, like area under a curve.
They have defined limits, unlike indefinite integrals, giving a specific numerical value based on the function and the limits of integration. When calculating definite integrals of a function over an interval \([-a, a]\), the result may vary depending on whether the function is odd, even, or neither. The notation \(\int_{-a}^{a} f(x) \, dx\) represents integrating the function \(f(x)\) from \(-a\) to \(a\). The integral aggregates the values of \(f(x)\) over the interval, producing a sum that corresponds to the net area between the curve and the x-axis.
When dealing with odd functions like \(\sin x\), we see unique properties emerge, helping simplify the calculation of these definite integrals.
Symmetric Property
The symmetric property of integrals can greatly simplify problems where functions exhibit symmetry.
If a function is odd, meaning \(f(-x) = -f(x)\) for all \(x\) in the domain, the integral over a symmetric interval \([-a, a]\) will always equal zero. To see why this is true, consider the integral breakdown:
  • The total integral \(\int_{-a}^{a} f(x) \, dx\) can be expressed as the sum of two integrals: \(\int_{-a}^{0} f(x) \, dx\) and \(\int_{0}^{a} f(x) \, dx\).

  • By substituting \(u = -x\) for the first integral, the limits change from \([-a, 0]\) to \([a, 0]\), and it can be rewritten in terms of \(u\): \(\int_{0}^{a} f(-u) \, du\).

  • Knowing \(f(-u) = -f(u)\) simplifies the expression to \(-\int_{0}^{a} f(u) \, du\).
Therefore, the symmetric property ensures that when these two integrals are added together, they cancel each other out, leading to a total sum of zero.
Integral Calculus
Integral calculus is a fundamental part of mathematics focusing on integration and computing areas under curves.
It encompasses techniques such as definite and indefinite integration, handling various types of functions, including odd and even. Understanding integral calculus involves:
  • Grasping the Fundamental Theorem of Calculus, which links differentiation and integration.

  • Applying various methods of integration like substitution and by parts.

  • Interpreting integrals as physical quantities such as area, volume, and average value.
Through the use of integral calculus, one can solve practical problems in engineering, physics, and other sciences.
It's crucial to identify the properties of the function being integrated, as they often illuminate pathways to simpler solutions.
As demonstrated in the exercise, recognizing the odd nature of a function allows the immediate conclusion of the result without cumbersome calculations.

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Most popular questions from this chapter

In Exercises \(75-78\) , let \(F(x)=\int_{a}^{x} f(t) d t\) for the specified function \(f\) and interval \([a, b] .\) Use a CAS to perform the following steps and answer the questions posed. \begin{equation} \begin{array}{l}{\text { a. Plot the functions } f \text { and } F \text { together over }[a, b] \text { . }} \\ {\text { b. Solve the equation } F^{\prime}(x)=0 . \text { What can you see to be true about }} \\ {\text { the graphs of } f \text { and } F \text { at points where } F^{\prime}(x)=0 \text { . Is your observation }} \\ {\text { borne out by Part } 1 \text { of the Fundamental Theorem coupled }} \\ {\text { with information provided by the first derivative? Explain your }} \\ {\text { answer. }}\\\\{\text { c. Over what intervals (approximately) is the function } F \text { increasing }} \\\ {\text { and decreasing? What is true about } f \text { over those intervals? }}\end{array} \end{equation} \begin{equation} \begin{array}{l}{\text { d. Calculate the derivative } f^{\prime} \text { and plot it together with } F . \text { What can }} \\ {\text { you see to be true about the graph of } F \text { at points where } f^{\prime}(x)=0 ?} \\\ {\text { Is your observation borne out by Part } 1 \text { of the Fundamental Theorem? }} \\ {\text { Explain your answer. }}\end{array} \end{equation} $$f(x)=2 x^{4}-17 x^{3}+46 x^{2}-43 x+12,\left[0, \frac{9}{2}\right]$$

Solve the initial value problems in Exercises \(55-60\) $$ \frac{d s}{d t}=8 \sin ^{2}\left(t+\frac{\pi}{12}\right), \quad s(0)=8 $$

Find the linearization of $$f(x)=2-\int_{2}^{x+1} \frac{9}{1+t} d t$$ at \(x=1.\)

The marginal cost of manufacturing \(x\) units of an electronic device is \(0.001 x^{2}-0.5 x+115 .\) If 600 units are produced, what is the production cost per unit?

Upper and lower sums for increasing functions $$ \begin{array}{l}{\text { a. Suppose the graph of a continuous function } f(x) \text { rises steadily }} \\ {\text { as } x \text { moves from left to right across an interval }[a, b] . \text { Let } P} \\ {\text { be a partition of }[a, b] \text { into } n \text { subintervals of equal length }} \\ {\Delta x=(b-a) / n . \text { Show by referring to the accompanying figure }}\\\ {\text { that the difference between the upper and lower sums for }} \\ {f \text { on this partition can be represented graphically as the area }} \\\ {\text { of a rectangle } R \text { whose dimensions are }[f(b)-f(a)] \text { by } \Delta x \text { . }} \\ {\text { (Hint: The difference } U-L \text { is the sum of areas of rectangles }}\\\\{\text { whose diagonals } Q_{0} Q_{1}, Q_{1} Q_{2}, \ldots, Q_{n-1} Q_{n} \text { lie approximately }} \\ {\text { along the curve. There is no overlapping when these rectan- }} \\ {\text { gles are shifted horizontally onto } R . \text { . }}\end{array} $$ $$ \begin{array}{l}{\text { b. Suppose that instead of being equal, the lengths } \Delta x_{k} \text { of the }} \\ {\text { subintervals of the partition of }[a, b] \text { vary in size. Show that }}\end{array} $$ $$ U-L \leq|f(b)-f(a)| \Delta x_{\max } $$ $$ \begin{array}{l}{\text { where } \Delta x_{\max } \text { is the norm of } P, \text { and hence that } \lim _{ \| P | \rightarrow 0}} \\\ {(U-L)=0}\end{array} $$
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