University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 5 - Section 5.3 - The Definite Integral - Exercises - Page 310: 68



Work Step by Step

Apply formula:$\int_{A}^{B} f(x)dx=\lim\limits_{n \to \infty} \Sigma_{k=1}^nf(a+k \triangle x)$ $\int_{-1}^{1} x^3 dx=\lim\limits_{n \to \infty}(\dfrac{2}{n}) \Sigma_{k=1}^n (-1+\dfrac{2k}{n})^3$ Thus, we have $\lim\limits_{n \to \infty}(\dfrac{2}{n})\Sigma_{k=1}^n(-1)+(\dfrac{8k^3}{n^3})+(\dfrac{6k}{n})-(\dfrac{12k^2}{n^2})=\lim\limits_{n \to \infty}(-\dfrac{2}{n})\Sigma_{k=1}^n(1)+(\dfrac{16}{n^4})\Sigma_{k=1}^nk^3+(\dfrac{12}{n^2})\Sigma_{k=1}^n k-(\dfrac{24}{n^3})\Sigma_{k=1}^nk^2$ This implies that $\lim\limits_{n \to \infty}[-2+4(1+\dfrac{1}{n})^2+6(1+\dfrac{1}{n})-4(1+\dfrac{1}{n})(2+\dfrac{1}{n})]=-2+4+6-8=0$
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