Calculus 8th Edition

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

Chapter 16 - Vector Calculus - 16.2 Line Integrals - 16.2 Exercises - Page 1124: 14



Work Step by Step

$I=\int_{1}^{4} (t)(\dfrac{dt}{2 \sqrt t})+t^2 dt+\sqrt t (2t) dt$ $=\int_{1}^{4} (\dfrac{t^{1/2}}{2})+t^2+2t^{3/2} dt$ $=(\dfrac{1}{2})(\dfrac{2}{3})t^{3/2}+\dfrac{t^3}{3}+\dfrac{4}{5}t^2\sqrt t]_1^{4}$ $=[\dfrac{8}{3}+\dfrac{64}{3}+\dfrac{128}{5}]-[\dfrac{1}{3}+\dfrac{1}{3}+\dfrac{4}{5}]$ $=\dfrac{70}{3}+\dfrac{124}{5}$ $=\dfrac{722}{15}$
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