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Chapter 5: Problem 114

a. Show that if \(f\) is odd on \([-a, a],\) then \(x\) $$\int_{-a}^{a} f(x) d x=0$$ 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 a symmetric interval is zero.

Step by step solution

01

Understanding Odd Functions

A function \(f(x)\) is called odd if it satisfies the condition \(f(-x) = -f(x)\) for all \(x\). This means that the function has point symmetry about the origin.
02

Setting Up the Integral

We need to evaluate the integral \(\int_{-a}^{a} f(x) \, dx\). Since \(f(x)\) is odd, \(f(-x) = -f(x)\) holds within the interval \([-a, a]\).
03

Symmetric Properties of the Integral

For an odd function \(f(x)\), the definite integral over a symmetric interval about the origin (like \([-a, a]\)) cancels out, because for every part of the area under \(f(x)\) from \(-a\) to \(0\), there is an equal and opposite area from \(0\) to \(a\). Therefore, \(\int_{-a}^{a} f(x) \, dx = 0\).
04

Testing with \(f(x)=\sin x\) and \(a=\pi/2\)

The function \(f(x) = \sin x\) is odd, because \(\sin(-x) = -\sin(x)\). We need to compute \(\int_{-\pi/2}^{\pi/2} \sin x \, dx\).
05

Evaluate the Integral

Calculate \(\int \sin x \, dx = -\cos x + C\). So, evaluate the integral from \(-\pi/2\) to \(\pi/2\):\[-\cos(\pi/2) + \cos(-\pi/2) = 0 - 0 = 0\].
06

Conclusion

Both theoretically, from the property of odd functions, and practically, by calculating the integral of \(\sin x\) over \([-\pi/2, \pi/2]\), we find \(\int_{-\pi/2}^{\pi/2} \sin x \, dx = 0\).

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Key Concepts

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

Odd Functions
An odd function is a mathematical concept that describes a certain type of symmetry in functions. For a function to be classified as odd, it must satisfy the equation \(f(-x) = -f(x)\) for every value of \(x\) in its domain. This means that if you plot the function on a coordinate system, it will exhibit symmetry about the origin.
This type of symmetry implies that for every point \((x, y)\) on the function, there is a corresponding point \((-x, -y)\).
  • Examples include functions like \(f(x) = \sin(x)\) and \(f(x) = x^3\).
  • These functions have a distinctive property when integrated over symmetric intervals, such as \([-a, a]\).
Understanding odd functions is crucial when dealing with integrals over symmetric intervals, particularly because they simplify calculations dramatically.
Symmetric Interval
A symmetric interval in the context of mathematics generally refers to an interval of the form \([-a, a]\), where \(a\) is a positive real number. Such intervals are perfectly balanced around zero, making them of particular interest when evaluating certain types of functions.
For odd functions, symmetric intervals play a key role in simplifying the evaluation of definite integrals.
  • The interval \([-a, a]\) ensures that every positive portion of the interval has a corresponding negative portion.
  • This balance is a foundation for understanding why the integral of an odd function over a symmetric interval results in zero.
Symmetric intervals are an important tool in calculus, especially when assessing the behavior of functions that are either odd or even.
Properties of Integrals
The properties of integrals are numerous, but when focusing on definite integrals involving odd functions over symmetric intervals, there's a specific property that stands out. It states that if you integrate an odd function over a symmetric interval, the result is always zero.
  • This is because the area under the curve from \(-a\) to 0 will cancel out with the area from 0 to \(a\).
  • For example, consider \(f(x) = \sin(x)\) over \([-\pi/2, \pi/2]\): the positive area under \(\sin(x)\) from 0 to \(\pi/2\) is exactly mirrored and canceled by the negative area from \(-\pi/2\) to 0.
Recognizing these properties allows for simpler computation in many integrals without having to perform complex calculations directly. This property is especially beneficial in both theoretical and practical applications of calculus.

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Most popular questions from this chapter

Use a CAS to perform the following steps: a. Plot the functions over the given interval. b. Partition the interval into \(n=100.200\), and 1000 subintervals of equal length, and evaluate the function at the midpoint of each subinterval. c. Compute the average value of the function values generated in part (b). d. Solve the equation \(f(x)=\) (average value) for \(x\) using the average value calculated in part (c) for the \(n=1000\) partitioning. $$f(x)=x \sin ^{2} \frac{1}{\pi} \quad \text { on } \quad\left[\frac{\pi}{4}, \pi\right]$$

Suppose that $$\int_{0}^{1} f(x) d x=3$$ Find $$\int_{-1}^{0} f(x) d x$$ if a. \(f\) is odd, \(\quad\) b. \(f\) is even.

Graph the function and find its average value over the given interval. $$f(x)=x^{2}-1 \text { on }[0, \sqrt{3}]$$

If \(f\) is a continuous function, find the value of the integral $$I=\int_{0}^{a} \frac{f(x) d x}{f(x)+f(a-x)}$$ by making the substitution \(u=a-x\) and adding the resulting integral to \(I\)

A basic property of definite integrals is their invariance under translation, as expressed by the equation $$\begin{aligned}&\int_{a}^{b} f(x) d x=\int_{a-c}^{b-c} f(x+c) d x\\\ &(1)\end{aligned}$$ The equation holds whenever \(f\) is integrable and defined for the necessary values of \(x .\) For example in the accompanying figure, show that $$\int_{-2}^{-1}(x+2)^{3} d x=\int_{0}^{1} x^{3} d x$$ because the areas of the shaded regions are congruent. (GRAPH CAN'T COPY). For each of the following functions, graph \(f(x)\) over \([a, b]\) and \(f(x+c)\) over \([a-c, b-c]\) to convince yourself that Equation (1) is reasonable. a. \(f(x)=x^{2}, \quad a=0, \quad b=1, \quad c=1\) b. \(f(x)=\sin x, \quad a=0, \quad b=\pi, \quad c=\pi / 2\) c. \(f(x)=\sqrt{x-4}, \quad a=4, \quad b=8, \quad c=5\)
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