Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 8 - Further Techniques and Applications of Integration - 8.2 Volume and Average Value - 8.2 Exercises - Page 440: 34

Answer

$38.83$

Work Step by Step

Use the formula for average value with a=0 and b=5. The average price is: $\frac{1}{5-0}\int^{5}_{0}(t(25-5t)+18)dx=\frac{1}{5-0}\int^{5}_{0}((25t-5t^{2})+18)dx=\frac{1}{5}(\frac{25t^{2}}{2}-\frac{5t^{3}}{3}+18t)|^{5}_{0} \approx \frac{233}{6} \approx 38.83$
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