Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 16: Integrals and Vector Fields - Practice Exercises - Page 1028: 6



Work Step by Step

We calculate the line integral by using substitution as follows: $$ x=1+9t \\ \implies dx=9 dt \\ y= 1+2t \\ \implies dy=2 \space dt \\ z=1+2\space t \\ \implies dz=2 \space dt $$ We need to plug the above values in the given integral. $$\int_{0}^{1} 9 \space dt-\sqrt{\dfrac{1+2t}{1+2t}}(2 dt) -\sqrt{\dfrac{1+2t}{1+2t} } \times (2 \space dt)\\=\int_{0}^{1}9 dt-2 dt -2dt=5\int_0^1 dt\\=5$$
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