Textbook Question
Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.
(a) If ƒ is symmetric about the line 𝓍 = 2 , then ∫₀⁴ ƒ(𝓍) d𝓍 = 2 ∫₀² ƒ(𝓍) d𝓍.
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8. Definite Integrals
Introduction to Definite Integrals
3:14 minutesProblem 8.3.63a
Textbook Question
63. Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.
a. If m is a positive integer, then ∫[0 to π] cos^(2m+1)(x) dx = 0.
Verified step by step guidance1
Step 1: Begin by analyzing the integral ∫[0 to π] cos^(2m+1)(x) dx. The integrand is cos^(2m+1)(x), which is an odd power of the cosine function. Recall that cosine is an even function, meaning cos(-x) = cos(x). However, raising cosine to an odd power makes the overall function odd, as odd powers of cosine reverse the sign when x is replaced with -x.
Step 2: Recall the property of definite integrals for odd functions. If f(x) is an odd function and the limits of integration are symmetric about zero (e.g., from -a to a), then ∫[-a to a] f(x) dx = 0. However, in this case, the limits of integration are from 0 to π, which are not symmetric about zero. This means the odd function property does not directly apply here.
Step 3: To determine whether the integral evaluates to zero, consider the behavior of cos^(2m+1)(x) over the interval [0, π]. The cosine function is positive on [0, π/2] and negative on [π/2, π]. Raising cosine to an odd power preserves the sign of the function, meaning cos^(2m+1)(x) is positive on [0, π/2] and negative on [π/2, π].
Step 4: Split the integral into two parts: ∫[0 to π/2] cos^(2m+1)(x) dx and ∫[π/2 to π] cos^(2m+1)(x) dx. The first integral represents the positive contribution, while the second integral represents the negative contribution. However, these contributions do not necessarily cancel out because the intervals [0, π/2] and [π/2, π] are not symmetric, and the magnitude of the function may differ across these intervals.
Step 5: Conclude that the statement 'If m is a positive integer, then ∫[0 to π] cos^(2m+1)(x) dx = 0' is false. Provide a counterexample by evaluating the integral for a specific value of m (e.g., m = 1) to show that the integral does not equal zero. This demonstrates that the odd power of cosine does not lead to cancellation over the interval [0, π].
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Definite Integrals
A definite integral represents the signed area under a curve between two specified limits. In this case, the integral of cos^(2m+1)(x) from 0 to π is evaluated to determine if it equals zero. Understanding how to compute definite integrals and the properties of the integrand is crucial for solving the problem.
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Properties of the Cosine Function
The cosine function is periodic and symmetric, specifically even, meaning cos(-x) = cos(x). For odd powers of cosine, such as cos^(2m+1)(x), the function exhibits symmetry about the y-axis, which can lead to cancellation of areas under the curve over symmetric intervals. This property is essential for determining the value of the integral.
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Odd and Even Functions
An odd function satisfies the condition f(-x) = -f(x), while an even function satisfies f(-x) = f(x). The function cos^(2m+1)(x) is odd because it is raised to an odd power. When integrating an odd function over a symmetric interval like [0, π], the result is zero, which is key to answering the question.
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