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: 21


$\approx 0.100001$ with an error of about $1.39 \times 10^{-11}$.

Work Step by Step

We integrate the integral with respect to $ x $ taking a lower limit as $0$ and upper limit as $0.4$. $\int_0^{0.1}\sqrt{1+x^4} dx=\int_0^{0.1} [1+\dfrac{x^4}{2}-\dfrac{x^{8}}{8}+...] dx $ or, $=0.1+\dfrac{(0.1)^5}{10}-\dfrac{(0.1)^9}{ 9 \cdot 8!}-....$ But $\dfrac{(0.1)^{9}}{9 \cdot 8} \approx 1.39 \times 10^{-11} $ and the proceeding term is greater than $10^{-8}$. So, the first two terms of the series would give the accuracy. $\int_0^{0.1}\sqrt{1+x^4} dx \approx 0.100001$ with an error of about $1.39 \times 10^{-11}$.
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