Calculus (3rd Edition)

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

Chapter 5 - The Integral - 5.2 The Definite Integral - Exercises - Page 245: 33



Work Step by Step

Let I= $\int_{0}^{4} (6t-3) dt$. Since $\int (6t-3) dt= \frac{6t^{2}}{2}-3t= 3t^{2}-3t=F(x)$, Therefore, by the second fundamental theorem, we get I= F(4)-F(0)= $(3\times4^{2}-3\times4)-0= 36$.
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