IB Mathematics SL 5.11 Definite integrals AA SL Paper 1- Exam Style Questions- New Syllabus
Question
Most-appropriate topic codes (IB Mathematics Analysis and Approaches SL):
• SL 5.11: Areas of a region enclosed by a curve \( y=f(x) \) and the x-axis; Areas between curves
• SL 2.9: Exponential functions and their graphs
• SL 5.10: Indefinite integral of \( e^x \)
▶️ Answer/Explanation
Understanding the region:
The curves are \( y = e^x \) (above) and \( y = -e^x \) (below), symmetric about the x-axis.
Vertical boundaries at \( x = -1 \) and \( x = 1 \).
METHOD 1: Using symmetry
The distance between curves at any \( x \) is: \( e^x – (-e^x) = 2e^x \)
Area = \( \int_{-1}^{1} [e^x – (-e^x)] \, dx = \int_{-1}^{1} 2e^x \, dx \)
\( = 2 \int_{-1}^{1} e^x \, dx \)
\( = 2 \left[ e^x \right]_{-1}^{1} \)
\( = 2(e^1 – e^{-1}) \)
\( = 2\left(e – \frac{1}{e}\right) \)
METHOD 2: Separate integrals
Area = \( \int_{-1}^{1} e^x \, dx – \int_{-1}^{1} (-e^x) \, dx \)
\( = \int_{-1}^{1} e^x \, dx + \int_{-1}^{1} e^x \, dx \)
\( = \left[ e^x \right]_{-1}^{1} + \left[ e^x \right]_{-1}^{1} \)
\( = (e^1 – e^{-1}) + (e^1 – e^{-1}) \)
\( = 2(e^1 – e^{-1}) \)
\( = 2\left(e – \frac{1}{e}\right) \)
METHOD 3: Recognizing symmetry about x-axis
Area = 2 × (Area above x-axis from \( x = -1 \) to \( x = 1 \))
\( = 2 \int_{-1}^{1} e^x \, dx \)
\( = 2(e^1 – e^{-1}) \)
\( = 2\left(e – \frac{1}{e}\right) \)
\( \boxed{2\left(e – \dfrac{1}{e}\right)} \) or \( \boxed{2e – \dfrac{2}{e}} \) or \( \boxed{2e – 2e^{-1}} \)