Question

Accumulated amount of money flow at time T. Assume that function gives the rate of flow of money in dollars per year over the given period, with continuous compounding at the given annual interest rate. Find the accumulated amount of money flow at the end of the time period.

f(t)=1000 + 200t, 10 years, 9%.

Answer 1

`A=30035.17 \$`

Explanation:

The accumulated amount of money flow equation is:

`A=e^(rT)\int^(T)_(0)f(t)e^(-rt)dt` (1)

Where:

• f(t) = 1000 + 200t (function)
• T = 10 years (period)
• r = 0.09 (annual interest rate.)

Then we have:

`A=e^(0.09\cdot 10)\int^(10)_(0)(1000 + 200t)e^(-0.09t)dt` (2)

Let's solve the integral. We can separate the functions and then use change variable and integration by parts.

`\int^(10)_(0)(1000+200t)e^(-0.09t)dt=\int^(10)_(0)1000e^(-0.09t)dt+\int^(10)_(0)  200te^(-0.09t)dt`

`\int^(10)_(0)(1000+200t)e^(-0.09t)dt=-11111.11e^(-0.09t)|^(10)_(0)+(e^(-0.09t) (-24691.4-2222.22t))|^(10)_(0)`

Now, we need to evaluate these integrals.

`\int^(10)_(0)(1000+200t)e^(-0.09t)dt=6593.67+5617.72=12211.39`

Finally the the accumulated amount of money flow will be:

Using (2),

`A=e^(0.09\cdot 10)12211.39=2.46*12211.39=30035.17 \$`

I hope it helps you!