Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 5: Integrals - Section 5.2 - Sigma Notation and Limits of Finite Sums - Exercises 5.2 - Page 266: 41



Work Step by Step

Consider $f(x)=1+x^2$ Here,we have $\Sigma_{i=1}^n (\dfrac{3}{n}) (1+k_i)^2=(\dfrac{27}{n^3}) \Sigma_{i=1}^n (3+i^2)$ This implies that $3+[(\dfrac{27n(n+1)(2n+1)}{6n^3})-(\dfrac{1}{n^3}) \Sigma_{i=1}^n i^2]=\dfrac{18+\dfrac{27}{n}+\dfrac{9}{n^2}}{2}+3$ Hence, $\lim\limits_{n \to \infty}\dfrac{18+\dfrac{27}{n}+\dfrac{9}{n^2}}{2}+3=9+3=12$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.