Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 4 - Integration - 4.5 Exercises - Page 301: 43

Answer

$ f_{(x)} = 2\cos{(\frac{x}{2})} + 4$

Work Step by Step

$f_{(x)}' = -\sin{(\frac{x}{2})} \longrightarrow f_{(x)} = \int [-\sin{(\frac{x}{2})}]\ dx = -\int \sin{(\frac{x}{2})}\ dx$ Let $u = \dfrac{x}{2} \longrightarrow du = \dfrac{dx}{2} \longrightarrow dx = 2du$ $-\int \sin{(\frac{x}{2})}\ dx = -\int \sin{(u)}\ 2\ du = -2\int \sin{(u)}\ du = -2(-\cos{(u)}) + C$ $\longrightarrow 2\cos{(u)} + C = 2\cos{(\frac{x}{2})} + C \longrightarrow f_{(x)} = 2\cos{(\frac{x}{2})} + C$ Since the graph passes through the point $(0, 6)$, then $f_{(0)} = 6 \longrightarrow 2\cos{(\frac{0}{2})} + C = 6 \longrightarrow 2\cos{0} + C = 6$ $\longrightarrow 2(1) + C = 6 \longrightarrow 2 + C = 6 \longrightarrow C = 4$ Hence $ f_{(x)} = 2\cos{(\frac{x}{2})} + 4$
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