Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 5 - The Integral - 5.4 The Fundamental Theorem of Calculus, Part I - Exercises - Page 258: 36



Work Step by Step

We have $$ \int_{0}^{3}|x^2-1| d x= \int_{0}^{1}|x^2-1| d x+\int_{1}^{3}|x^2-1| d x\\ = -\int_{0}^{1}x^2-1\ d x+\int_{1}^{3}x^2-1\ d x\\{4}x^4|_{0}^{3}\\ =-(\frac{1}{3}x^3 +x)_{0}^{1}+(\frac{1}{3}x^3 +x)_{1}^{3}\\ =-\frac{1}{3}-1+\frac{27}{3}+1-\frac{1}{3}-1=\frac{22}{3}. $$
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