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Chapter 14: Problem 43

Investments If you invest $$\$ 10,000$$ at \(8 \%\) interest compounded continuously, what is the average amount in your account over one year?

Short Answer

Expert verified
The average amount in the account over one year when investing \$10,000 at 8% interest compounded continuously is approximately \(\$10,410.87\).

Step by step solution

01

Formula for continuous compounding interest

We will use the formula for continuous compounding interest, which is given by: \( A(t) = P \cdot e^{(r \cdot t)} \) Where: - \(A(t)\) represents the amount in the account at time \(t\) - \(P\) is the principal amount (initial investment) - \(r\) is the interest rate expressed as a decimal - \(e\) is the base of the natural logarithm - \(t\) is the time in years
02

Apply the parameters given in the exercise

We know that: - \(P = \$10,000\) - \(r = 8\% = 0.08\) - \(t\) ranges from 0 to 1 year So, the formula for the amount in the account at any time during the first year is: \( A(t) = 10000 \cdot e^{(0.08 \cdot t)} \)
03

Find the average amount in the account

To find the average account balance over one year, we will integrate the balance function with respect to time over the interval \([0,1]\) (one year) and divide the result by 1 year. Average amount in the account over one year = \( \frac{1}{1} \int_{0}^{1} 10000 \cdot e^{(0.08 \cdot t)} dt \)
04

Calculation

Now let's calculate the integral: \( \int_{0}^{1} 10000 \cdot e^{(0.08 \cdot t)} dt = \frac{10000}{0.08} (e^{0.08 \cdot t}) \Big |_0^1 \) \( = \frac{10000}{0.08} (e^{0.08 \cdot 1}) - \frac{10000}{0.08} (e^{0.08 \cdot 0}) \) \( = \frac{10000}{0.08} (e^{0.08} - e^{0}) \)
05

Evaluate the value of the average amount

Plug in the value of \(e^{0.08}\) and simplify: \( \frac{10000}{0.08} (1.083287 - 1) = \frac{10000}{0.08} \cdot 0.083287 \) \( \approx \$10,410.87 \) So, the average amount in the account over the first year is approximately \(\$10,410.87\).

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