Answer
The accumulated amount after 3 years is:
\[
A \approx 63748.43
\]
Work Step by Step
To find the accumulated amount of an investment with a continuous money flow rate of \( f(t) = 20,000 \) per year for $3$ years at an interest rate of $4%$ compounded continuously, we can use the concept of continuous compounding.
The formula for the future value of a continuous money flow \( f(t) \) compounded continuously at an interest rate \( r \) over a period \( T \) is given by:
\[
A = \int_{0}^{T} f(t) e^{r(T-t)} \, dt
\]
Here:
- \( f(t) = 20,000 \)
- \( r = 0.04 \)
- \( T = 3 \)
Substitute \( f(t) \) into the integral:
\[
A = \int_{0}^{3} 20000 e^{0.04(3-t)} \, dt
\]
Simplify the exponent:
\[
A = \int_{0}^{3} 20000 e^{0.12 - 0.04t} \, dt
\]
Factor out the constant \( e^{0.12} \):
\[
A = 20000 e^{0.12} \int_{0}^{3} e^{-0.04t} \, dt
\]
Now, integrate \( e^{-0.04t} \):
\[
\int e^{-0.04t} \, dt = \frac{e^{-0.04t}}{-0.04} = -25 e^{-0.04t}
\]
Evaluate the definite integral from $0$ to $3$:
\[
A = 20000 e^{0.12} \left[ -25 e^{-0.04t} \right]_{0}^{3}
\]
Calculate the integral at the bounds:
\[
A = 20000 e^{0.12} \left( -25 e^{-0.04 \cdot 3} + 25 e^{-0.04 \cdot 0} \right)
\]
Simplify the exponentials:
\[
A = 20000 e^{0.12} \left( -25 e^{-0.12} + 25 \right)
\]
Factor out \( -25 \):
\[
A = 20000 e^{0.12} \cdot -25 \left( e^{-0.12} - 1 \right)
\]
Simplify using \( e^{0.12} e^{-0.12} = 1 \):
\[
A = -20000 \cdot 25 \left( 1 - e^{0.12} \right)
\]
Simplify the constant multiplication:
\[
A = -500000 \left( 1 - e^{0.12} \right)
\]
Calculating the value we get,
\[
A=63748.42579
\]
Therefore, the accumulated amount after 3 years is:
\[
A \approx 63748.43
\]
So, the accumulated amount at an interest rate of 4% compounded continuously is approximately $63,748.43$.
