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Part 3. In general, if f(x)dx = c then, if f is an even function then, | f(x)dx = 2f(x)a -a if f is an odd function then, f(x)dx =

Part 3. In general, if f(x)dx = c then, if f is an even function then, | f(x)dx = 2f(x)a -a if f is an odd function then, f(x)dx =

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Part1. The functionf is an even function if f(-x) = f(x) for all x in the domain of f. The graph of an even function is symmetric with respect to the y-axis. The functionf is an odd function if f(-x) = -f(x) for all x in the domain of f. The graph of án odd function is symmetric with respect to the origin. Part 2. Now, let's consider the definite integral of an even and odd function over a symmetric interval by pictorially examining the area under the curve. -9 -8 -5 -2 -1 1294 5. 6 8. -1 -2 -3 -4 -5 -6 -7 -8 -9 This is the graph of f(x) = x² (which is an even function) with the area under the curve from x = -3 to x = 3 shaded in blue. Notice that the amount of shaded area from x = -3 to x = 0 and from x = 0 to x = 3 is equal. We therefore conclude that %3D 3 3 2dx = 2. 2dx
Transcribed Image Text:Part1. The functionf is an even function if f(-x) = f(x) for all x in the domain of f. The graph of an even function is symmetric with respect to the y-axis. The functionf is an odd function if f(-x) = -f(x) for all x in the domain of f. The graph of án odd function is symmetric with respect to the origin. Part 2. Now, let's consider the definite integral of an even and odd function over a symmetric interval by pictorially examining the area under the curve. -9 -8 -5 -2 -1 1294 5. 6 8. -1 -2 -3 -4 -5 -6 -7 -8 -9 This is the graph of f(x) = x² (which is an even function) with the area under the curve from x = -3 to x = 3 shaded in blue. Notice that the amount of shaded area from x = -3 to x = 0 and from x = 0 to x = 3 is equal. We therefore conclude that %3D 3 3 2dx = 2. 2dx
-9 -8 -7 -8 -5 -4 -3 1 4 -1 -8 -9 This is the graph of f(x) = x' (which is an odd function) with the area under the curve from x = -2 to x = 2 shaded in blue. Notice that the amount of shaded area from x = -2 to x = 0 and from x = 0 to x = 2 is equal. However, the the definite integral will produce a negative value for the area to the left of x = 0 for this function. We therefore conclude that x dx = - x² dx 0. x' dx + x² dx = 0 -2 x' dx = 0 -2 Part 3. In general, a if f(x)dx = c then, a if f is an even function then, | f(x)dx 2f(x)a -a if f is an odd function then, f(x)dx = %3D
Transcribed Image Text:-9 -8 -7 -8 -5 -4 -3 1 4 -1 -8 -9 This is the graph of f(x) = x' (which is an odd function) with the area under the curve from x = -2 to x = 2 shaded in blue. Notice that the amount of shaded area from x = -2 to x = 0 and from x = 0 to x = 2 is equal. However, the the definite integral will produce a negative value for the area to the left of x = 0 for this function. We therefore conclude that x dx = - x² dx 0. x' dx + x² dx = 0 -2 x' dx = 0 -2 Part 3. In general, a if f(x)dx = c then, a if f is an even function then, | f(x)dx 2f(x)a -a if f is an odd function then, f(x)dx = %3D
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