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 1126: 41



Work Step by Step

Work done$W=\int_C F\cdot dr=\int_0^{1} \lt2t-t^2,-t^2 +3t-1, 1-t-4t^2 \gt \cdot \lt 2,1,-1 \gt dt$ $=\int_0^{1} 2(2t-t^2)+(-t^2 +3t-1)-( 1-t-4t^2) dt$ $=\int_0^{1} 4t-2t^2-t^2 +3t-1- 1+t+4t^2 dt$ $=\int_0^{1} t^2+8t-2dt$ $=[\dfrac{t^3}{3}+4t^2-2t]_0^1$ $=[\dfrac{1^3}{3}+4(1^2-0)-2(1-0)]_0^1$ $=\dfrac{1}{3}+4-2$ $=\dfrac{7}{3}$
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