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| rev | Admin | (Created page with "'''Solution: C''' <math display="block"> \begin{aligned} & 1.05=1+i=v^{-1}=e^\delta \text { so } \delta=\ln (1.05) \text { and } e^{\delta t}=1.05^t \\ & P V=\int_0^n f(t) v^t d t=\int_0^{10}(1000+100 t) e^{-\delta t} d t \end{aligned} </math> Thus after 10 years, the accumulated value is <math display="block"> (1.05)^{10} P V=1.05^{10} \int_0^{10}(1000+100 t) e^{-\delta t} d t=1.05^{10} \int_0^{10}(1000+100 t)\left(1.05^{-t}\right) d t </math> '''References'''...") | Nov 26'23 at 18:19 | +698 |