Applications of the Definite Integral

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Future Value of a continuous income stream Formula :

∫^N₀ K$e^{r(N-t)}$dt

K = dollars per year

N = years at interest rate r

r = interest rate

Example 1 :

At an annual rate of $1000, money is being deposited into a savings account everyday. The savings account pays 6% interest that is compounded continuously. What is approximately the amount of money in the account after 5 years?

Solution :

∫⁵₀$1000e^{0.06(5-t)}$dt

=${\frac {1000}{-0.06}}e^{0.06(5-t)}$ |⁵₀

=${\frac {1000}{-0.06}}(1-e^{0.3})$

= approximately = 5831

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