Answer
$$\int^1_0(8s^3-12s^2+5)ds=3$$
Work Step by Step
$$A=\int^1_0(8s^3-12s^2+5)ds$$ $$A=\Big(\frac{8s^4}{4}-\frac{12s^3}{3}+5s\Big)\Big]^1_0$$ $$A=\Big(2s^4-4s^3+5s\Big)\Big]^1_0$$ $$A=(2-4+5)-0$$ $$A=3$$
University Calculus: Early Transcendentals (3rd Edition)
by Hass, Joel R.; Weir, Maurice D.; Thomas Jr., George B.
Published by
Pearson
ISBN 10:
0321999584
ISBN 13:
978-0-32199-958-0
Chapter 5 - Practice Exercises - Page 343: 74
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