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To solve this problem, we need to compare the population growth under two different conditions between the year 2010 and the year 2015.

1. **Determine the time span**: 
   - \( t = 0 \) corresponds to the year 2010.
   - \( t = 5 \) corresponds to the year 2015.

2. **Population growth rate without policy changes**: 
   - Given \( P_1'(t) = 5e^{0.03t} \) thousand people per year.
   - To find the population at any time \( t \), we need to integrate \( P_1'(t) \):
     \[
     P_1(t) = \int 5e^{0.03t} \, dt = \frac{5}{0.03} e^{0.03t} + C = \frac{500}{3} e^{0.03t} + C
     \]
   - At \( t = 0 \), \( P_1(0) \) (the population in 2010) can be taken as \( P_0 \). So,
     \[
     P_1(0) = \frac{500}{3} + C = P_0
     \]
   - Therefore,
     \[
     C = P_0 - \frac{500}{3}
     \]
   - So,
     \[
     P_1(t) = \frac{500}{3} e^{0.03t} + P_0 - \frac{500}{3}
     \]
   - At \( t = 5 \):
     \[
     P_1(5) = \frac{500}{3} e^{0.15} + P_0 - \frac{500}{3}
     \]

3. **Population growth rate with policy changes**:
   - Given \( P_2'(t) = 5 + 0.03t + 0.003t^2 \) thousand people per year.
   - To find the population at any time \( t \), we need to integrate \( P_2'(t) \):
     \[
     P_2(t) = \int (5 + 0.03t + 0.003t^2) \, dt = 5t + 0.03 \frac{t^2}{2} + 0.003 \frac{t^3}{3} + C = 5t + 0.015t^2 + 0.001t^3 + C
     \]
   - At \( t = 0 \), \( P_2(0) \) (the population in 2010) can also be taken as \( P_0 \). So,
     \[
     P_2(0) = C = P_0
     \]
   - Therefore,
     \[
     P_2(t) = 5t + 0.015t^2 + 0.001t^3 + P_0
     \]
   - At \( t = 5 \):
     \[
     P_2(5) = 5 \cdot 5 + 0.015 \cdot 5^2 + 0.001 \cdot 5^3 + P_0 = 25 + 0.375 + 0.125 + P_0 = 25.5 + P_0
     \]

4. **Difference in population growth between the two scenarios**:
   - Calculate the population in 2015 without policy changes:
     \[
     P_1(5) = \frac{500}{3} e^{0.15} + P_0 - \frac{500}{3}
     \]
   - Calculate the population in 2015 with policy changes:
     \[
     P_2(5) = 25.5 + P_0
     \]
   - Subtract \( P_2(5) \) from \( P_1(5) \) to find the difference:
     \[
     \Delta P = P_1(5) - P_2(5) = \left(\frac{500}{3} e^{0.15} + P_0 - \frac{500}{3}\right) - (25.5 + P_0)
     \]
     \[
     \Delta P = \frac{500}{3} (e^{0.15} - 1) - 25.5
     \]

Let's compute the exact value:

\[
e^{0.15} \approx 1.1618
\]
\[
\frac{500}{3} \approx 166.67
\]
\[
\Delta P \approx 166.67 \cdot (1.1618 - 1) - 25.5
\]
\[
\Delta P \approx 166.67 \cdot 0.1618 - 25.5
\]
\[
\Delta P \approx 26.98 - 25.5
\]
\[
\Delta P \approx 1.48
\]

Therefore, the population in 2015 with the new policies would be approximately 1.48 thousand people less than it would be without the new policies.

Do you want details or have any questions? Here are some questions you might ask:

1. How do you integrate exponential functions?
2. Can you explain the method for solving definite integrals?
3. How do you handle population growth models in calculus?
4. What are the applications of continuous compounding in real-world scenarios?
5. Can you show examples of different population growth models?

**Tip:** When solving problems involving exponential functions, remember to simplify your calculations by handling constants and exponentials separately.

Learn how to solve population growth models using calculus techniques. This problem involves comparing population growth under different conditions between 2010 and 2015, utilizing exponential functions and integration methods. Suitable for undergraduate students studying calculus and population dynamics.

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