Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 5 - Section 5.3 - The Fundamental Theorem of Calculus - 5.3 Exercises: 36

Answer

$6\sqrt 6$ - $2\sqrt 3$

Work Step by Step

$\int^{18}_{1} \sqrt \frac{3}{z} dz$ The first step is to simplify the integral and bring the constant outside the integral: $= \int^{18}_{1} \frac{\sqrt 3}{\sqrt z} dz$ $= \sqrt 3 \int^{18}_{1} \frac{1}{\sqrt z} dz$ $= \sqrt 3 \int^{18}_{1} z^{-1/2} dz$ Next step is to solve for the definite integral using power rule. [ $\int^b_a x^{n} dx = (\frac{x^{n+1}}{n+1})|^b_a$ ] $= \sqrt{3} (\frac{z^{\frac{1}{2}}}{\frac{1}{2}})|^{18}_1$ $= \sqrt 3 (2\sqrt{z})|^{18}_1$ Plug in the values from 18 to 1. $ = \sqrt{3}( 2\sqrt{18} - 2\sqrt{1} )$ Simplify until final answer: $= \sqrt{3} (6\sqrt{2} - 2)$ $= 6\sqrt{6} - 2\sqrt{3}$
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