Problem 1 · Single-variable calculus
A landscaped area is bounded by the curves y = x² and y = 4x, with x and y in meters. The area enclosed between the two curves is most nearly:
- A.10.7 m²
- B.21.3 m²
- C.32.0 m²
- D.53.3 m²
Show the answer and solution
Answer: A: 10.7 m²
The curves meet at x = 0 and x = 4. Area = ∫₀⁴ (4x − x²) dx = [2x² − x³/3]₀⁴ = 32 − 21.33 = 10.7 m². Adding the two integrals (32 + 21.33 = 53.3 m²) counts the region under the parabola instead of removing it.
Single-variable calculus · Reference Handbook: Mathematics
