Revision as of 23:02, 17 November 2023 by Admin (Created page with "'''Solution: E''' The accumulated value to time 4 is <math display = "block"> \int_{1}^{3}100e^{0.5t}e^{0.88(4-t)}d t=100e^{0.32}\int_{1}^{3}e^{0.42t}d t=100e^{0.32}\frac{e^{0.42t}}{0.42}\biggl|_{1}^{3}=\frac{100e^{0.32}(e^{1.26}-e^{0.42})}{0.42}=657. </math> {{soacopyright | 2023 }}")
Exercise
ABy Admin
Nov 17'23
Answer
Solution: E
The accumulated value to time 4 is
[[math]]
\int_{1}^{3}100e^{0.5t}e^{0.88(4-t)}d t=100e^{0.32}\int_{1}^{3}e^{0.42t}d t=100e^{0.32}\frac{e^{0.42t}}{0.42}\biggl|_{1}^{3}=\frac{100e^{0.32}(e^{1.26}-e^{0.42})}{0.42}=657.
[[/math]]
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