IBDP Maths SL 5.11 Definite integrals AA HL Paper 1- Exam Style Questions- New Syllabus
Question
Determine the total area enclosed by the curves \( y = e^x \) and \( y = -e^x \), between the vertical lines \( x = -1 \) and \( x = 1 \).
Most-appropriate topic codes (IB Mathematics Analysis and Approaches 2021):
• SL 5.11: Definite integrals; areas of a region enclosed by a curve and the x-axis (including areas between curves) — whole question
• SL 5.10: Indefinite integrals of elementary functions, including \( e^x \) — integration step
• SL 2.9: Exponential functions \( f(x) = a^x \) and \( f(x) = e^x \); graphs and properties — understanding the curves
• SL 5.10: Indefinite integrals of elementary functions, including \( e^x \) — integration step
• SL 2.9: Exponential functions \( f(x) = a^x \) and \( f(x) = e^x \); graphs and properties — understanding the curves
▶️ Answer/Explanation
Step 1 — Describe the region:
The curve \( y = e^x \) lies above the x-axis and the curve \( y = -e^x \) lies below it. For \( -1 \le x \le 1 \), the vertical distance between the two curves is
\( e^x – (-e^x) = 2e^x \).
Step 2 — Form the definite integral:
Area \( = \int_{-1}^{1} \big[ e^x – (-e^x) \big] \, dx = \int_{-1}^{1} 2e^x \, dx \).
Step 3 — Integrate:
\( \int 2e^x \, dx = 2e^x \).
Step 4 — Apply the limits:
Area \( = \left[ 2e^x \right]_{-1}^{1} = 2e – 2e^{-1} \).
Final answer (exact form):
\( \boxed{2e – \frac{2}{e}} \), equivalently \( \boxed{2\left(e – \frac{1}{e}\right)} \).