Answer:
a) [tex]P= $56,972.5[/tex]
b) [tex]A=$1,26,792.3[/tex]
Step-by-step explanation:
Given Data:
Interest rate=[tex]r= 0.08[/tex] per year
No. of years=[tex]t=10[/tex]
Rate of continuous money flow is given by the function
[tex]f(t)=2000[/tex]
a) to find the present value of money
[tex]P=\int\limits^n_0 {f(t)e^{-rt} } \, dt[/tex]
Put f(t)=2000 and n=10 years and r=0.08
[tex]P=\int\limits^n_0 {2000e^{-0.08t} } \, dt[/tex]
Now integrate
[tex]P= {2000(\frac{e^{-0.08t}}{-0.08} )[/tex]
[tex]P= -\frac{2000}{0.08} (e^{-0.08*10}-e^{-0.08*0})[/tex]
[tex]P= -\frac{2000}{0.08} (e^{-0.8}-e^{0})[/tex]
[tex]P= -\frac{2000}{0.08} (0.4493-2.7282)[/tex]
[tex]P= -\frac{2000}{0.08} (-2.2789)[/tex]
[tex]P= -25000(-2.2789)[/tex]
[tex]P= $56972.5[/tex]
(b) to find the accumulated amount of money at t=10
[tex]A=P(e^{rt} )[/tex]
Where P is the present worth already calculated in part a
[tex]A=56972.5(e^{0.08*10} )[/tex]
[tex]A=56972.5(e^{0.8} )[/tex]
[tex]A=56972.5(2.2255 )[/tex]
[tex]A=$1,26,792.3[/tex]
