I cannot figure this out for the life of me. Maybe I am overthinking it.

Money is transferred into an account at the rate of R(t)=5000t(e^-0.6t)

If the account pays 2% interest compounded continuously, how much will accumulate in the account over a 9-year period?

Money at 2% grows like e^0.02t

The amount deposited at time t gets to grow for (9-t) years. Seems to me like the account balance after 9 years is thus

∫[0,9] R(t) * e^(.02(9-t)) dt

I may have misspoken. Take a look here for a discussion of the topic.

faculty.atu.edu/mfinan/2243/business65.pdf

actually, it appears I was right after all...

To solve this problem, we need to use the formula for continuous compound interest and integrate the given transfer rate function.

The continuous compound interest formula is given by:

A = P * e^(rt)

Where:
A = the final amount accumulated in the account
P = the initial amount of money transferred into the account
r = the interest rate (in decimal form)
t = the time period in years

In this case, the initial amount transferred is given by the transfer rate function R(t) = 5000t(e^(-0.6t)). To find the total amount accumulated, we need to integrate R(t) from 0 to 9 years.

∫(from 0 to 9) [5000t(e^(-0.6t))] dt

To solve the integral, we can use integration by parts. The integration by parts formula is:

∫ u dv = uv - ∫ v du

Let's assign u = t and dv = 5000t(e^(-0.6t)) dt. Then, calculate du and v:

du = dt
v = ∫ 5000t(e^(-0.6t)) dt = -8333.33(e^(-0.6t)) - 13888.89(e^(-0.6t)) + 13888.89

Now, we can apply the integration by parts formula:

∫ t * [5000t(e^(-0.6t))] dt = t * (-8333.33(e^(-0.6t)) - 13888.89(e^(-0.6t)) + 13888.89) - ∫ (-8333.33(e^(-0.6t)) - 13888.89(e^(-0.6t)) + 13888.89) dt

Simplifying, we get:

A = -8333.33te^(-0.6t) - 13888.89te^(-0.6t) + 13888.89t + c^(-0.6t) + C

Note that c is a constant from the integration process, and C is another constant from the integration, called the constant of integration.

To find the total amount accumulated over a 9-year period, we need to evaluate A at t = 9 and subtract the initial amount P (which is 0), since there was no money in the account initially:

Amount accumulated = A(9) - P = [-8333.33(9)e^(-0.6*9) - 13888.89(9)e^(-0.6*9) + 13888.89(9) + c^(-0.6*9)] - 0

Now, you can evaluate the expression using a calculator or software to find the amount accumulated in the account over the 9-year period.