The present value $P V$ of a capital asset that provides a perpetual stream of revenue that flows continuously at a rate of $R(t)$ dollars per year is given by $\newline$ $P V=\int_{0}^{\infty} R(t) e^{-r t} d t$ $\newline$ where $r$ , expressed as a decimal, is the annual rate of interest compounded continuously. $\newline$ (a) Find the present value of the asset if it provides a constant return of $\$ 110$ per year and $r=9 \%$ . $\newline$ (Use decimal notation. Give your answer to the nearest dollar.) $\newline$ $P V=$ $\newline$ dollars $\newline$ (b) Find the present value of the asset if it provides a return of $R(t)=1100+70 t$ dollars per year and $r=6 \%$ . $\newline$ (Use decimal notation. Give your answer to the nearest dollar.) $\newline$ $P V=$ $\newline$ dollars

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Grade 8

Evaluate a linear function: word problems

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Q. The present value

P V

of a capital asset that provides a perpetual stream of revenue that flows continuously at a rate of

R(t)

dollars per year is given by

\newline

P V=\int_{0}^{\infty} R(t) e^{-r t} d t

\newline

where

r

, expressed as a decimal, is the annual rate of interest compounded continuously.

\newline

(a) Find the present value of the asset if it provides a constant return of

\$ 110

per year and

r=9 \%

\newline

(Use decimal notation. Give your answer to the nearest dollar.)

\newline

P V=

\newline

dollars

\newline

(b) Find the present value of the asset if it provides a return of

R(t)=1100+70 t

dollars per year and

r=6 \%

\newline

(Use decimal notation. Give your answer to the nearest dollar.)

\newline

P V=

\newline

dollars

Calculate Constant Return PV: Let's solve part (a) of the problem where the asset provides a constant return of $\$110$ per year and the annual interest rate $r$ is $9\%$ or $0.09$ in decimal form. $\newline$ The present value (PV) formula for a constant return is: $\newline$ $PV = \int_{0}^{\infty} R(t) * e^{(-rt)} dt$ $\newline$ Since $R(t)$ is constant at $\$110$ per year, the equation simplifies to: $\newline$ $PV = \int_{0}^{\infty} 110 * e^{(-0.09t)} dt$
Solve Integral for Constant Return: To solve the integral, we use the formula for the integral of an exponential function: $\newline$ $\int e^{at} dt = \frac{1}{a} \cdot e^{at} + C$ $\newline$ Applying this to our integral, we get: $\newline$ $PV = 110 \cdot \int_{0}^{\infty} e^{-0.09t} dt$ $\newline$ = $110$ \cdot \left[\left(-\frac{ $1$ }{ $0$ . $09$ }\right) \cdot e^{ $-0$ . $09$ t}\right]_{ $0$ }^{\infty}
Evaluate Integral Limits: Now we evaluate the expression at the limits of integration: $\newline$ PV = \(110 \times \left[\left(-\frac{ $1$ }{ $0$ . $09$ }\right) \times e^{( $-0$ . $09$ \times \infty)} - \left(-\frac{ $1$ }{ $0$ . $09$ }\right) \times e^{( $-0$ . $09$ \times $0$ )}\right] $\newline$ = $110$ \times \left[ $0$ - \left(-\frac{ $1$ }{ $0$ . $09$ }\right)\right] $\newline$ = $110$ \times \left(\frac{ $1$ }{ $0$ . $09$ }\right)
Find Present Value: Simplifying the expression, we find the present value: $\newline$ $PV = 110 \times \left(\frac{1}{0.09}\right)$ $\newline$ $= 110 \times 11.1111\ldots$ $\newline$ $\approx 1222.22$
Calculate Variable Return PV: Rounding to the nearest dollar, the present value for part (a) is: $\newline$ $PV \approx \$1222$
Split Integral for Variable Return: Now let's solve part (b) where the return is given by $R(t) = 1100 + 70t$ dollars per year and the annual interest rate $r$ is $6\%$ or $0.06$ in decimal form. $\newline$ The present value (PV) formula for a variable return is the same: $\newline$ $PV = \int_{0}^{\infty} R(t) \cdot e^{(-rt)} dt$ $\newline$ Substituting $R(t)$ with $1100 + 70t$ , we get: $\newline$ $PV = \int_{0}^{\infty} (1100 + 70t) \cdot e^{(-0.06t)} dt$
Solve Integral $1$ : This integral is more complex because it involves both a constant and a linear term. We can split the integral into two parts: $\newline$ $PV = \int_{0}^{\infty} 1100 \cdot e^{-0.06t} \, dt + \int_{0}^{\infty} 70t \cdot e^{-0.06t} \, dt$
Apply Integration by Parts: The first integral is similar to the one we solved in part (a), so we can solve it in the same way: $\newline$ $PV1 = 1100 \times \int_{0}^{\infty} e^{(-0.06t)} dt$ $\newline$ = $1100 \times \left[\frac{-1}{0.06} \times e^{(-0.06t)}\right]$ from $0$ to $\infty$ $\newline$ = $1100 \times \left(\frac{1}{0.06}\right)$ $\newline$ $\approx 18333.33$
Evaluate Integral $2$ : The second integral requires integration by parts, which is given by: $\newline$ $\int u \, dv = uv - \int v \, du$ $\newline$ Let's choose $u = t$ and $dv = 70 \cdot e^{(-0.06t)} \, dt$ . Then $du = dt$ and $v = \frac{-1}{0.06} \cdot e^{(-0.06t)}$ .
Add Present Value Parts: Applying integration by parts, we get: $\newline$ $PV_2 = \int_{0}^{\infty} t \cdot 70 \cdot e^{(-0.06t)} \, dt$ $\newline$ = $70 \cdot \left[ t \cdot \left(-\frac{1}{0.06}\right) \cdot e^{(-0.06t)} - \int \left(-\frac{1}{0.06}\right) \cdot e^{(-0.06t)} \, dt\right]_{0}^{\infty}$
Round to Nearest Dollar: Evaluating the integral and the limits, we find: $\newline$ PV2 = 70 \times [0 - 0 - ((-1/0.06) \times (-1/0.06) \times e^{(-0.06 \times 0)} - 0)]$\newline = 70 \times [(1/0.06)^2] $\newline$ \approx 19444.44$
Round to Nearest Dollar: Evaluating the integral and the limits, we find: $\newline$ PV2 = 70 \times [0 - 0 - ((-1/0.06) \times (-1/0.06) \times e^{(-0.06 \times 0)} - 0)]$\newline = 70 \times [(1/0.06)^2] $\newline$ \approx 19444.44$Adding the two parts of the present value together: $\newline$ PV = PV1 + PV2$\newline \approx 18333.33 + 19444.44 $\newline$ \approx 37777.77$
Round to Nearest Dollar: Evaluating the integral and the limits, we find: $\newline$ PV2 = 70 \times [0 - 0 - ((-1/0.06) \times (-1/0.06) \times e^{(-0.06 \times 0)} - 0)]$\newline = 70 \times [(1/0.06)^2] $\newline$ \approx 19444.44$Adding the two parts of the present value together: $\newline$ PV = PV1 + PV2$\newline \approx 18333.33 + 19444.44 $\newline$ \approx 37777.77$Rounding to the nearest dollar, the present value for part (b) is: $\newline$ $PV \approx \$(37778)$