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# If income is continuously collected at a rate of $f(t)$ dollars per year and will be invested at a constant interest rate $r$ (compounded continuously) for a period of $T$ years, then the future value of the income is given by $\int_0^T f(t) e^{r (T - t)}\ dt$. Compute the future value after 6 years for income received at a rate of $f(t) = 8000 e^{0.04t}$ dollars per year and invested at $6.2 \%$ interest.

## $\$ 65,230.48$#### Topics Applications of Integration ### Discussion You must be signed in to discuss. ##### Top Calculus 2 / BC Educators ##### Calculus 2 / BC Bootcamp Lectures Join Bootcamp ### Watch More Solved Questions in Chapter 8 ### Video Transcript Okay, so the question gives us the following formula. The integral from zero to capital t of F 50 times e to the R times, the quantity capital, T minus T d t is the future value of income, and then it also gives us half of T R and Capital T and asked us to compete. So, using the formula they gave us, we get integral from 0 to 6 of 1000 e 0.4 t times e to 0.62 times six minus t duty. And this doing one simplification, we're going to get taking out constant terms. And then distributing the exponential is together and combining the light terms, we have this constant factor of e 0.37 to allow. Then everything that's left is a exponential factor left and the integral. And so now we just divide by the coefficient of tea and then keep the exponential term the same and evaluate so writing and out we have 8000 e to 0.372 and you're going to divide at the front was going to be negative. Then we have 0.0 to 2 Times e to minus 0.0 to 2 t taken from 0 to 6. And so when we do that, we have approximately minus 527,000 52.9 times and then plugging in the six and zero get minus 0.124 which finally gives us a future value of approximately 65,000$230 and 48 cents.

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