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: 34



Work Step by Step

Let I= $\int^{2}_{-3}(4x+7)dx$ Since $\int(4x+7)dx=$$\frac{4x^{2}}{2}+7x= 2x^{2}+7x=F(x)$ Therefore, by the second fundamental theorem, we get $I= F(2)-F(-3)= $$(2\times2^{2}+7\times2)-[2\times(-3^{2})+(7\times(-3))]$ =22-(-3)= 25
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