Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 4 - Integrals - 4.3 The Fundamental Theorem of Calculus - 4.3 Exercises - Page 328: 36


$6\sqrt6 -2\sqrt3$

Work Step by Step

Evaluate the Integral: $\int^{18}_{1}\sqrt{\frac{3}{z}} dz $ Recall the 2nd part of the Fundamental Theorem of Calculus: $∫^b_af(x)dx=F(b)−F(a)$ Find $F(x)$: $F(x) = \int\sqrt{\frac{3}{x}} dx$ $F(x) = \int\frac{\sqrt3}{\sqrt x}dx$ $F(x) = \sqrt3 \int\frac{1}{\sqrt x}dx$ $F(x) = \sqrt3 \int x^{-\frac{1}{2}}dx$ $F(x) = \sqrt3 \frac{x^{(-1/2) +1}}{(-1/2) +1}$ $F(x) = \sqrt3(2\sqrt x)$ Now Evaluate: $F(b) - F(a)$ $F(18) - F(1)$ $\sqrt3(2\sqrt{18}) - \sqrt3(2\sqrt1)$ $\sqrt3(6\sqrt2 - 2)$ $6\sqrt6 -2\sqrt3$
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