Calculus (3rd Edition)

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

Chapter 5 - The Integral - 5.5 The Fundamental Theorem of Calculus, Part II - Exercises - Page 262: 12


$$f(\theta) =76-\frac{5}{2} \sin^2\theta -9\sin \theta. $$

Work Step by Step

We have $$f(\theta)=\int_{\sin \theta}^{4}5t+9 \ dt =\frac{5}{2}t^2 +9t|_{\sin \theta}^{4}\\ =40-\frac{5}{2} \sin^2\theta+36-9\sin \theta\\ =76-\frac{5}{2} \sin^2\theta -9\sin \theta. $$
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