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: 43



Work Step by Step

According to Rule (4) in Table 5.4 : $\int_{0}^{2} (2t-3)dt=\int_{0}^{2} 2tdt+\int_{0}^{2} (-3)dt$ According to Rule (3) in Table 5.4 and because $0\lt2$, by Equation (2): $\int_{0}^{2} 2tdt+\int_{0}^{2} (-3)dt=2\int_{0}^{2} tdt-3\int_{0}^{2} 1dt=2\cdot(\frac{2^2}{2}-\frac{0^2}{2})-3\cdot(2-0)=2\cdot(2-0)-3\cdot2=4-6=-2$
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