IB Mathematics SL 5.11 Definite integrals AA SL Paper 2- Exam Style Questions- New Syllabus
Question
Most-appropriate topic codes (Mathematics: analysis and approaches guide):
SL 5.11: Definite integrals and analytical approach: \( \int_a^b g'(x)dx = g(b) – g(a) \) — Method 1
SL 5.5: Anti-differentiation with a boundary condition to determine the constant term — Method 2
▶️ Answer/Explanation
Method 1 – Using a definite integral
The start of 2026 is \( t = 4 \) years after the start of 2022.
The change in population \( \Delta P \) from \( t = 0 \) to \( t = 4 \) is:
\( \Delta P = \int_{0}^{4} \frac{dP}{dt} \, dt = \int_{0}^{4} -104000 e^{-0.0145t} \, dt \)
Integrating using the rule for \( e^{at} \):
\( \Delta P = \left[ \frac{-104000}{-0.0145} e^{-0.0145t} \right]_{0}^{4} \)
\( \Delta P = \left[ 7172413.793 e^{-0.0145t} \right]_{0}^{4} \)
\( \Delta P = 7172413.793(e^{-0.058} – 1) \)
\( \Delta P \approx -404165.8 \)
Final population \( P(4) = P(0) + \Delta P \):
\( P(4) = 6.78 \times 10^6 – 404165.8 \approx 6375834.2 \)
Answer: \( \boxed{6.38 \times 10^6} \) (to 3 s.f.)
Method 2 – Using indefinite integration
Integrate the differential equation to find the general form of \( P(t) \):
\( P(t) = \int -104000 e^{-0.0145t} \, dt \)
\( P(t) = 7172413.793 e^{-0.0145t} + C \)
Apply the initial condition \( P(0) = 6.78 \times 10^6 \):
\( 6780000 = 7172413.793(e^{0}) + C \)
\( C = 6780000 – 7172413.793 = -392413.793 \)
The population function is \( P(t) = 7172413.793 e^{-0.0145t} – 392413.793 \).
At \( t = 4 \):
\( P(4) = 7172413.793 e^{-0.058} – 392413.793 \approx 6375834.2 \)
Answer: \( \boxed{6.38 \times 10^6} \) (to 3 s.f.)