Calculus Concepts: An Informal Approach to the Mathematics of Change 5th Edition

Published by Brooks Cole
ISBN 10: 1-43904-957-2
ISBN 13: 978-1-43904-957-0

Chapter 6 - Analyzing Accumulated Change: Integrals in Action - 6.1 Activities - Page 423: 22

Answer

Converges to $125$

Work Step by Step

Let us consider that $I_n=\int_{0}^{n} 5x e^{-0.02x^2} \ dx$ Suppose $a=-0.02x^2 \implies da=-0.04x \ dx$ $I_n=-125\int_{0}^{-0.02n^2} e^{a} \ da=125 (1-e^{-0.02n^2})$ We define the given integral as a limit of $I_n$ as: $I=\lim\limits_{n \to \infty} I_n=\int_{0}^{\infty} 5x e^{-0.02x^2} \ dx \\=\lim\limits_{n \to \infty} 125 (1-e^{-0.02n^2})=125$ This means that the limit exist and the given integral converges to $125$.
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