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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...

Edit: I forgot to add part c, I made an edit and added it.

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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 $P_T$.

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  • $\begingroup$ Thank you so much! I understood the question now, that was very helpful. $\endgroup$
    – Dewton
    Sep 12 '19 at 21:33
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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: $$P_0 = P_0 \cdot e^{rT}-P_0$$ $$2 = e^{0.08\cdot T}$$ $$\ln{2} = 0.08\cdot T$$ $$T = \frac{\ln{2}}{0.08}$$

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  • $\begingroup$ Thank you so much :), that was very helpful! $\endgroup$
    – Dewton
    Sep 12 '19 at 21:33
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You need to add the initial principal $P_0$ back to get the result

$$P_{T} = P_{0}e^{rT}$$

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