A fund earns interest at a rate equivalent to 5% per annum effective. A person makes continuous deposits to the Fund for 10 years The rate at which deposits are made is 1000+100t per annum at time t (in years). How much will the person have in the fund at the end of the ten year period. (Set up the proper integral but do not evaluate. The letters i, v, δ should not appear in your integral.)

• [[math]]\int_0^{10}(1000+100 t)e^{10-t} d t [[/math]]
• [[math]] \int_0^{10}(1000+100 t)\left(1.05^{t}\right) d t [[/math]]
• [[math]]1.05^{10} \int_0^{10}\exp{(1000+100 t)\left(1.05^{-t}\right)} d t [[/math]]
• [[math]]1.05^{10} \int_0^{10}(1000+100 t)\left(1.05^{t}\right) d t [[/math]]
• [[math]]1.05^{10} \int_0^{10}(1000+100 t)\left(1.05^{-t}\right) d t [[/math]]

References

Hlynka, Myron. "University of Windsor Old Tests 62-392 Theory of Interest". web2.uwindsor.ca. Retrieved November 23, 2023.

Solution: C

[[math]] \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]] (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

Hlynka, Myron. "University of Windsor Old Tests 62-392 Theory of Interest". web2.uwindsor.ca. Retrieved November 23, 2023.