# Applications of Integration, Continuous Compound Interest

I need some help in this question, I do not think I understood it completely, I will show my attempt of solutions.

The question is:

A bank may compound interest over various lengths of time: yearly, half-yearly, quarterly, monthly, daily, and so on. If interest is compounded instantaneously, we can show that $$I=\int_{0}^{T} P_{0}re^{rt}\; dt$$ where $$I$$ is the interest accrued, $$P_{0}$$ is the initial investment, $$r$$ is the rate of interest per annum as a decimal, and $$T$$ is the period of the loan in years.

a) Show that the amount of money in an account at time $$T$$ is given by $$P_{T} = P_{0}e^{rT}$$

b) How long will it take for an amount to double at a rate of 8% p.a.?

c) A block of land was bought for $$55$$ dollars in 1940 and sold for $$196000$$ dollars in 2007 at the same time of the year. What rate of interest, compounded instantaneously, would produce this increase in the same time?

My attempt to solve part a: \begin{align*} I = \int_{0}^{T} P_{0}re^{rt} \; dt &= \left[P_{0}e^{rt} \right]_{0}^{T}\\ &=\left[P_{0}e^{rT}-P_{0}e^{0} \right]\\ &=P_{0}e^{rT}-P_{0} \end{align*}

however, it is not as $$P_{T}=P_{0}e^{rT}$$. I don't think my attempt was correct. I think this is an equation for the continuous compound interest, but I don't know how to arrive to it. As for part b, I know that $$r=0.08$$ but I don't know how only knowing this will help me to obtain $$T$$.I would really appreciate some help...

This issue is because the integral you calculated is equal to $$I_T$$, not $$P_T$$. If $$I_T$$ is the interest at time $$T$$, $$P_0$$ is the principal, and $$P_T$$ is the total amount at time $$T$$, then $$P_T=P_0+I_T$$ This checks out, since you calculated the integral $$I_T=P_0 e^{rT}-P_0$$, implying that $$P_T=P_0+P_0 e^{rT}-P_0=P_0 e^{rT}$$ as claimed. So there is no problem with your solution; you just forgot that the integral represented $$I_T$$ instead of $$P_T$$.
As for the second part, you simply wish for the $$I$$ to be equal $$P_0$$ (because your profit will be equal to the money you invested in the first place). You've already calculated the integral, so we get: $$P_0 = P_0 \cdot e^{rT}-P_0$$ $$2 = e^{0.08\cdot T}$$ $$\ln{2} = 0.08\cdot T$$ $$T = \frac{\ln{2}}{0.08}$$
You need to add the initial principal $$P_0$$ back to get the result
$$P_{T} = P_{0}e^{rT}$$