University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 5 - Section 5.4 - The Fundamental Theorem of Calculus - Exercises - Page 321: 73



Work Step by Step

Here, we have $\dfrac{dc}{dx}=\dfrac{1}{2 \sqrt x}$ $\dfrac{dc}{dx}=(1/2)x^{-1/2}$ This implies that $c=\int_0^{x} \dfrac{1}{2} t dt$ or, $[t^{1/2}]_0^{x}=\sqrt x$ Thus, $c(100)-c(1)=\sqrt {100}-\sqrt 1=10-1=9$
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