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: 67



Work Step by Step

Apply formula:$\int_{A}^{B} f(x)dx=\lim\limits_{n \to \infty} \Sigma_{k=1}^nf(a+k \triangle x)$ $\int_{-1}^{2} 3x^2-2x+1 dx=\lim\limits_{n \to \infty}(\dfrac{3}{n}) \Sigma_{k=1}^n 3(-1+\dfrac{3k}{n})^2-2 (-1+\dfrac{3k}{n})+1$ Thus, we have $\lim\limits_{n \to \infty}(\dfrac{18}{n})\Sigma_{k=1}^n1+(\dfrac{18}{n^3})\Sigma_{k=1}^nk^2-(\dfrac{72}{n^2})\Sigma_{k=1}^n k=(\dfrac{18}{n})(n)+(\dfrac{18}{n^3})(\dfrac{n(n+1)(2n+1)}{6})-(\dfrac{72}{n^2}) (\dfrac{n(n+1)}{2})$ This implies that $18+(27/2)(1+\dfrac{1}{n})(2+\dfrac{1}{n})-36(1+\dfrac{1}{n})=18+27-36=9$
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