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 - Chapter Review - Review Exercises - Page 456: 58

Answer

An oil leak from an uncapped well is polluting a bay at a rate of $ f(t)=125 e^{-0.025 t}$ gallons per year. The total amount of oil that will enter the bay, assuming the well is never capped, is given by: $$ \begin{aligned} \int_{0}^{\infty} 125 e^{-0.025 t} d t &=\lim _{b \rightarrow \infty} \int_{0}^{b} 125 e^{-0.025 t} d t \\ &=5000 \text{ gallons} \end{aligned} $$

Work Step by Step

An oil leak from an uncapped well is polluting a bay at a rate of $ f(t)=125 e^{-0.025 t}$ gallons per year. The total amount of oil that will enter the bay, assuming the well is never capped, is given by: $$ \begin{aligned} \int_{0}^{\infty} 125 e^{-0.025 t} d t &=\lim _{b \rightarrow \infty} \int_{0}^{b} 125 e^{-0.025 t} d t \\ &=\left.\lim _{b \rightarrow \infty}\left(\frac{125}{-0.025} e^{-0.025 t}\right)\right|_{0} ^{b}\\ &=\lim _{b \rightarrow \infty}(-5000)\left(e^{-0.025 b}-e^{0}\right)\\ &=-5000\left(\lim _{b \rightarrow \infty} \frac{1}{e^{0.025 b}}-1\right)\\ &=-5000\left((0)-1\right)\\ &=5000. \end{aligned} $$ The total amount of oil that will enter the bay $5000$ gallons.
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