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Ch. 8 - Integration Techniques
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 8, Problem 8.2.82e

82. A family of exponentials The curves y = x * e^(-a * x) are shown in the figure for a = 1, 2, and 3.


Graph of y = x * e^(-a * x) showing curves for a = 1, 2, and 3, illustrating a family of exponential decay functions.


e. Does this pattern continue? Is it true that A(1, ln b) = a² * A(a, (ln b)/a)?

Verified step by step guidance
1
First, understand the function given: \(y = x e^{-a x}\), where \(a\) is a parameter that changes the shape of the curve. The graph shows this function for \(a = 1, 2, 3\).
Next, identify what \(A(a, b)\) represents in the problem. Typically, \(A(a, b)\) might denote the value of the function or an area related to the function at specific points. Here, it likely means the value of the function \(y = x e^{-a x}\) evaluated at \(x = b\) for a given \(a\), so \(A(a, b) = b e^{-a b}\).
To check the given relation \(A(1, \ln b) = a^2 \cdot A(a, \frac{\ln b}{a})\), substitute the definitions: - Compute \(A(1, \ln b) = (\ln b) e^{-1 \cdot \ln b}\). - Compute \(A(a, \frac{\ln b}{a}) = \left(\frac{\ln b}{a}\right) e^{-a \cdot \frac{\ln b}{a}}\).
Simplify the exponentials using the property \(e^{\ln x} = x\) and \(e^{-\ln x} = \frac{1}{x}\): - \(e^{-\ln b} = \frac{1}{b}\), - \(e^{-a \cdot \frac{\ln b}{a}} = e^{-\ln b} = \frac{1}{b}\).
After simplification, compare both sides of the equation: - Left side: \(A(1, \ln b) = (\ln b) \cdot \frac{1}{b} = \frac{\ln b}{b}\), - Right side: \(a^2 \cdot \left( \frac{\ln b}{a} \cdot \frac{1}{b} \right) = a^2 \cdot \frac{\ln b}{a b} = a \cdot \frac{\ln b}{b}\). Since the left side is \(\frac{\ln b}{b}\) and the right side is \(a \cdot \frac{\ln b}{b}\), the equality holds only if \(a = 1\). Therefore, the pattern does not continue for all \(a\).

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