Question

If the birth rate of a population is
b(t) = 2500e0.024t
people per year and the death rate is
d(t)= 1400e0.018t people per year,
find the area between these curves for 0 ≤ t ≤ 10. (Round your answer to the nearest integer.)

Answer 1

To find the area between the birth rate curve \( b(t) \) and the death rate curve \( d(t) \) over the interval \( 0 \leq t \leq 10 \), we first need to set up the integral of the difference between the two functions over this interval.

The functions are given as:
- Birth rate:
\[
b(t) = 2500 e^{0.024t}
\]
- Death rate:
\[
d(t) = 1400 e^{0.018t}
\]

The area \( A \) between the two curves from \( t = 0 \) to \( t = 10 \) is given by the integral of their difference:
\[
A = \int_0^{10} (b(t) - d(t)) \, dt = \int_0^{10} (2500 e^{0.024t} - 1400 e^{0.018t}) \, dt
\]

We will calculate this integral step by step.

### Step 1: Calculate the integral

Compute the integral:
\[
A = \int_0^{10} (2500 e^{0.024t} - 1400 e^{0.018t}) \, dt
\]

This can be split into two separate integrals:
\[
A = \int_0^{10} 2500 e^{0.024t} \, dt - \int_0^{10} 1400 e^{0.018t} \, dt
\]

#### Integrating \( 2500 e^{0.024t} \)

The integral of \( e^{kt} \) is \( \frac{1}{k} e^{kt} \). Thus, we have:
\[
\int 2500 e^{0.024t} \, dt = \frac{2500}{0.024} e^{0.024t} = \frac{2500}{0.024} e^{0.024t}
\]

Evaluate from 0 to 10:
\[
= \left. \frac{2500}{0.024} e^{0.024t} \right|_0^{10} = \frac{2500}{0.024} \left( e^{0.24} - 1 \right)
\]

#### Integrating \( 1400 e^{0.018t} \)

Similarly,
\[
\int 1400 e^{0.018t} \, dt = \frac{1400}{0.018} e^{0.018t}
\]

Evaluate from 0 to 10:
\[
= \left. \frac{1400}{0.018} e^{0.018t} \right|_0^{10} = \frac{1400}{0.018} \left( e^{0.18} - 1 \right)
\]

### Step 2: Putting it all together

Now substitute back into the area expression:
\[
A = \frac{2500}{0.024} \left(e^{0.24} - 1\right) - \frac{1400}{0.018} \left(e^{0.18} - 1\right)
\]

### Step 3: Calculate numerical values

1. Calculate \( e^{0.24} \) and \( e^{0.18} \):
- \( e^{0.24} \approx 1.271 \)
- \( e^{0.18} \approx 1.197 \)

2. Compute the area:
\[
A \approx \frac{2500}{0.024}(1.271 - 1) - \frac{1400}{0.018}(1.197 - 1)
\]
\[
\approx \frac{2500}{0.024} \cdot 0.271 - \frac{1400}{0.018} \cdot 0.197
\]
\[
\approx 104166.67 \cdot 0.271 - 77777.78 \cdot 0.197
\]
\[
\approx 28219.44 - 15370.50 \approx 12848.94
\]

### Result

Rounding to the nearest integer, the area between the curves over the interval \( 0 \leq t \leq 10 \) is approximately:
\[
\boxed{12849}
\]