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 309: 30



Work Step by Step

Because $0.5\lt2.5$, by Equation (2): $\int_{0.5}^{2.5} x dx=\frac{(2.5)^2}{2}-\frac{(0.5)^2}{2}=\frac{6.25}{2}-\frac{0.25}{2}=\frac{6}{2}=3$.
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