University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 9 - Section 9.10 - The Binomial Series and Applications of Taylor Series - Exercises - Page 549: 18


$$\approx 0.3546472$$ with an error of about $10^{-5}$.

Work Step by Step

We integrate the integral with respect to $ x $ and take a lower limit as $0$ and upper limit as $0.35$. $\int_0^{0.35}\sqrt [3] {1+x^2} dx=\int_0^{0.35} [1+\dfrac{x^2}{3}-\dfrac{x^{4}}{9}+...] dx $ or, $=0.35-\dfrac{(0.35)^3}{9}-\dfrac{(0.35)^5}{45}+\dfrac{5(0.35)^7}{567}....$ But $\dfrac{5(0.35)^7}{567} \approx 5.67 \times 10^{-6} $ and the proceeding term is greater than $10^{-5}$. So, the first three terms of the series would give the accuracy. $\int_0^{0.35}\sqrt [3] {1+x^2} dx \approx 0.3546472$ with an error of about $10^{-5}$.
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