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

1000 e^{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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