find the future value at 5.25% interest, compounded continuously for 6 years, of the continuous income…

Answer:

$5037$

Explanation:

Step1: Recall the formula for future - value of a continuous income stream

The formula for the future - value $F$ of a continuous income stream with a rate of flow $f(t)$ compounded continuously at an interest rate $r$ for $T$ years is $F = e^{rT}\int_{0}^{T}f(t)e^{-rt}dt$. Here, $r = 0.0525$, $T = 6$, and $f(t)=725e^{-0.02t}$. So, $F = e^{0.0525\times6}\int_{0}^{6}725e^{-0.02t}e^{-0.0525t}dt$.

Step2: Simplify the integrand

Using the property of exponents $a^m\times a^n=a^{m + n}$, we have $e^{-0.02t}e^{-0.0525t}=e^{-(0.02t + 0.0525t)}=e^{-0.0725t}$. So, $F = e^{0.315}\times725\int_{0}^{6}e^{-0.0725t}dt$.

Step3: Integrate $e^{-0.0725t}$

The antiderivative of $e^{-0.0725t}$ is $\frac{e^{-0.0725t}}{- 0.0725}$. Evaluating $\int_{0}^{6}e^{-0.0725t}dt=\left[-\frac{e^{-0.0725t}}{0.0725}\right]{0}^{6}=-\frac{e^{-0.0725\times6}}{0.0725}+\frac{e^{0}}{0.0725}=-\frac{e^{-0.435}}{0.0725}+\frac{1}{0.0725}$. $e^{-0.435}\approx0.6472$, so $\int{0}^{6}e^{-0.0725t}dt=\frac{1 - 0.6472}{0.0725}=\frac{0.3528}{0.0725}\approx4.8662$.

Step4: Calculate $e^{0.315}$

$e^{0.315}\approx1.3707$.

Step5: Calculate the future - value $F$

$F = 1.3707\times725\times4.8662$. $1.3707\times725 = 1.3707\times(700 + 25)=1.3707\times700+1.3707\times25=959.49+34.2675 = 993.7575$. $F=993.7575\times4.8662\approx5037$.