## Continuous Exponential:

In continuous compounded interest, we will often see the value of the Euler's number {eq}e {/eq} which is approximately 2.71828. This is derived from the fact that if evaluate the limit {eq}\lim _{x\to \:\infty }\left(1+\frac{1}{x}\right)^x = e {/eq}, which derives from the idea of compounded not every minute nor second but everytime, that is, continuosly.

We want to get the area under the curve of the function of the profit. Given that the profit is growing continuously, then by the equation of continuous compounding

\begin{align} P&=P_0e^{rt} \\ P_1 &= 15.3e^{r_1t} \\ P_1&=15.3e^{\left(\frac{3.5}{100}\right)t} \\ P_2 &= 15.3e^{r_2t}dt \\ P_2&=15.3e^{\left(\frac{4}{100}\right)t} \\ \end{align}

Now, we have {eq}P_0 = 15.3 {/eq} million, then the cumulative difference would be the integral of the difference of the two annual rates

\begin{align} P&=\int _0^4P_2-P_1dt \\ P&=\int _0^415.3e^{r_1t}-15.3e^{r_2t}dt \\ P&=\int _0^415.3e^{\left(\frac{4}{100}\right)t}-15.3e^{\left(\frac{3.5}{100}\right)t}dt \\ \end{align}

Performing the integration

\begin{align} P&=\int _0^415.3e^{\left(\frac{4}{100}\right)t}-15.3e^{\left(\frac{3.5}{100}\right)t}dt \\ P&=\int _0^415.3e^{\left(\frac{4}{100}\right)t}\times \frac{\frac{4}{100}}{\frac{4}{100}}dt-15.3e^{\left(\frac{3.5}{100}\right)t}\times \frac{\frac{3.5}{100}}{\frac{3.5}{100}}dt \\ P&=\int _0^4\frac{15.3}{\frac{4}{100}}e^{\left(\frac{4}{100}\right)t}\times \frac{4}{100}dt-\frac{15.3}{\frac{3.5}{100}}e^{\left(\frac{3.5}{100}\right)t}\times \frac{3.5}{100}dt \\ P&=\left[\frac{15.3}{\frac{4}{100}}e^{\left(\frac{4}{100}\right)t}-\frac{15.3}{\frac{3.5}{100}}e^{\left(\frac{3.5}{100}\right)t}\right] \Biggr|_0^4 \\ P&=\left[\left(\frac{15.3}{\frac{4}{100}}e^{\left(\frac{4}{100}\right)\left(4\right)}-\frac{15.3}{\frac{3.5}{100}}e^{\left(\frac{3.5}{100}\right)\left(4\right)}\right)-\left(\frac{15.3}{\frac{4}{100}}e^{\left(\frac{4}{100}\right)\left(0\right)}-\frac{15.3}{\frac{3.5}{100}}e^{\left(\frac{3.5}{100}\right)\left(0\right)}\right)\right] \\ P&=\left[\left(\frac{15.3}{\frac{4}{100}}e^{\left(\frac{4}{100}\right)\left(4\right)}-\frac{15.3}{\frac{3.5}{100}}e^{\left(\frac{3.5}{100}\right)\left(4\right)}\right)-\left(\frac{15.3}{\frac{4}{100}}\left(1\right)-\frac{15.3}{\frac{3.5}{100}}\left(1\right)\right)\right] \\ P &\approx \left[\left(-53.96606\right)-\left(-54.64285\right)\right] \\ P &\approx 0.67679 \\ \end{align}

$$\boxed { P \approx \ 0.68 \hspace{1mm} million }$$