Respuesta :
Answer:
Step-by-step explanation:
a)
Using formular for present value [tex]P=\int\limits^t_0 {f(x)e^{rx}} \, dx \\\\f(x)=5000e^{-0.01x}[/tex] at [tex]r=0.08[/tex] and [tex]t=8[/tex]
Present value =[tex] \int\limits^8_0 {5000e^{-0.01x}e^{-0.08x}} \, dx\\\\\int\limits^8_0 {5000e^{-0.09x}} \, dx[/tex]
Taking the constants outside, we get
[tex]=5000\int\limits^8_0 {e^{-0.09x}} \, dx ...(1)[/tex]
Integrating [tex]\int {e^{-0.09x}} \, dx[/tex] we get
[tex]=(\frac{5000}{-0.09}e^{-0.09x})\limits^8_0[/tex]
as
[tex]\int {e^{ax}}dx=\frac{e^{ax}}{a}\\\\=\frac{5000e^{-0.72}}{-0.09}+\frac{5000}{0.09}\\\\=$28,513.76[/tex]
Hence, the present value is $28,513.76
b)
Using formular for final amount [tex]e^{rt}\int\limits^t_0 {f(x)e^{-rx}}dx[/tex] at [tex]r=0.08, t=8\\\\=e^{0.08(8)}\int\limits^8_0 {5000e^{-0.01x}e^{0.08x}}dxx[/tex]
Takig the constant outside-
[tex]=5000e^{0.64}\int\limits^8_0 {e^{-0.09x}}dx[/tex]
Integrating [tex]\int {e^{-0.09x}}dx[/tex], we get
[tex]=e^{0.64}\frac{5000}{-0.09}(e^{-0.09x})\limits^8_0[/tex]
as
[tex]\int {e^{ax}}dx=\frac{e^{ax}}{a}\\\\=e^{0.64}(28,513.76)\\=$54,075.81[/tex]
Hence, final value is $54,075.81
The present value is 28,513.76 and the final amount is 54,075.81 at an interest rate of 8% compounded continuously.
Given that,
Money Flow The rate of a continuous flow of money starts at $5000 and decreases exponentially at 1% per year for 8 years.
We have to determine,
The present value and final amount at an interest rate of 8% compounded continuously.
According to the question,
Money Flow The rate of a continuous flow of money starts at $5000 and decreases exponentially at 1% per year for 8 years.
The exponential function represents the present value is,
[tex]P =\int\limits^t_0 {f(x) . e^{rx}} \, dx[/tex]
Using the formula for present value [tex]f(x) = 5000e^{-0.01x}[/tex] at and r = 0,08 and t = 8 years.
Substitute the values in the formula,
[tex]P =\int\limits^8_0 {5000e^{-0.01} . e^{-0.08x}} \, dx\\\\ p = \int\limits^8_0 {5000e^{-0.09} } \, dx\\[/tex]
Integrating both sides,
[tex]P = 5000\int\limits^8_0 {e^{-0.09} } \, dx\\\\P = [\dfrac{5000}{-0.09} \times e^{-0.09x}]^x_0\\\\P = \dfrac{5000e^{-0.72}}{-0.09} + \dfrac{5000}{0.09}\\\\P = 28,513.76[/tex]
The present value is 28,513.76.
And the final value is,
Using the formula for present value [tex]f(x) = 5000e^{-0.01x}[/tex] at and r = 0,08 and t = 8 years.
Substitute the values in the formula,
[tex]P =\int\limits^8_0 {5000e^{-0.01} . e^{-0.08x}} \, dx\\\\ p = \int\limits^8_0 {5000. e^{0.64}e^{-0.09} } \, dx\\[/tex]
Integrating both sides,
[tex]P = 5000. e^{0.64}\int\limits^8_0 {e^{-0.09} } \, dx\\\\P = [\dfrace^{e^0.64}{5000}{-0.09} \times e^{-0.09x}]^x_0\\\\P = e^{-0.64} \times \dfrac{5000e^{-0.72}}{-0.09} + \dfrac{5000}{0.09}\\\\P = 54,075.81[/tex]
The final value is 54,075.81.
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