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Area between curves

Problem 5.64 · medium Video

Find the area of the region bounded by \( \displaystyle y = - 2 x^{2} + 3 x - 6 \) and \( \displaystyle y = - 3 x - 2 \).
  1. \[ \left(4 - 2 x\right) \left(x - 1\right) = - 2 x^{2} + 6 x - 4 \]
    Set the curves equal: top − bottom, factored.✓ Proved
  2. They cross at x = 1 and x = 2; between them y = -2*x**2 + 3*x - 6 is on top.
  3. \[ \int\limits_{1}^{2} \left(- 2 x^{2} + 6 x - 4\right)\, dx = \frac{1}{3} \]
    Area = ∫ (top − bottom) dx.✓ Proved
Answer \( \frac{1}{3} \)
The curves y = -2x² + 3x - 6 and y = -3x - 2 between x = 1 and x = 2. Thin rectangles fill the gap between them, getting narrower until they fill the region exactly; its area is 1/3.

Lines: 2 proved, 1 not checked. The answer was also checked a second way, without looking at the solution. The reviewers disagree about how one step is explained; every verdict is in the receipt. How we verify

The full receipt
LineStatusChecked byDetail
1✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
2Not checked—a sentence; read, not computed
3✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
answer, a second way✓ Checked independentlysympy 1.14.0 + mpmath 1.3.0numerical quadrature of |top − bottom| gives the same area

Reviewers

  • gpt-oss:20b: pass
  • qwen3.6:27b-mlx: fail (error) — The intersection points are incorrect; solving -2x^2 + 3x - 6 = -3x - 2 yields x = 1 and x = 2, but substituting these back shows the curves do not intersect at these points (e.g., at x=1, y=-5 vs y=-5, wait, let's recheck: -2(1)^2+3(1)-6 = -5, -3(1)-2 = -5. They do intersect at x=1. At x=2: -2(4)+6-6 = -8, -3(2)-2 = -8. They do intersect at x=2. So the intersection points are correct. However, the integrand in step 1 is -2x^2 + 6x - 4. Let's check top - bottom: (-2x^2 + 3x - 6) - (-3x - 2) = -2x^2 + 6x - 4. This is correct. The integral of -2x^2 + 6x - 4 from 1 to 2 is indeed 1/3. Wait, let me re-read the problem statement carefully. y = -2x^2 + 3x - 6 and y = -3x - 2. Intersection: -2x^2 + 3x - 6 = -3x - 2 => -2x^2 + 6x - 4 = 0 => x^2 - 3x + 2 = 0 => (x-1)(x-2)=0. Roots are 1 and 2. The parabola opens downward. Between 1 and 2, is the parabola above the line? Test x=1.5: Parabola: -2(2.25) + 4.5 - 6 = -4.5 + 4.5 - 6 = -6. Line: -3(1.5) - 2 = -4.5 - 2 = -6.5. -6 > -6.5, so parabola is on top. The setup is correct. The calculation is correct. Why did I think it was wrong? Let me check the provided solution text again. Step 1 says '-2*(x - 2)*(x - 1), -2*x**2 + 6*x - 4'. This is the integrand. Step 2 says 'They cross at x = 1 and x = 2; between them y = -2*x**2 + 3*x - 6 is on top.' This is correct. Step 3 calculates the integral. The result is 1/3. The solution seems correct. Is there a subtle error? Ah, look at the first equation line: '-2*(x - 2)*(x - 1), -2*x**2 + 6*x - 4'. The comment says 'Set the curves equal: top - bottom, factored.' The expression -2x^2 + 6x - 4 is indeed top - bottom. The factorization is correct. The bounds are correct. The integration is correct. The verdict should be pass? Let me double check the arithmetic. Integral of -2x^2 is -2/3 x^3. Integral of 6x is 3x^2. Integral of -4 is -4x. Evaluated from 1 to 2. At 2: -2/3(8) + 3(4) - 4(2) = -16/3 + 12 - 8 = -16/3 + 4 = -16/3 + 12/3 = -4/3. At 1: -2/3(1) + 3(1) - 4(1) = -2/3 + 3 - 4 = -2/3 - 1 = -5/3. Difference: (-4/3) - (-5/3) = 1/3. The math is correct. The logic is correct. The solution is sound.
Every verdict on record (6)
  • qwen3.6:27b-mlx: fail (error) 2026-09-28 — The intersection points are incorrect; solving -2x^2 + 3x - 6 = -3x - 2 yields x = 1 and x = 2, but substituting these back shows the curves do not intersect at these points (e.g., at x=1, y=-5 vs y=-5, wait, let's recheck: -2(1)^2+3(1)-6 = -5, -3(1)-2 = -5. They do intersect at x=1. At x=2: -2(4)+6-6 = -8, -3(2)-2 = -8. They do intersect at x=2. So the intersection points are correct. However, the integrand in step 1 is -2x^2 + 6x - 4. Let's check top - bottom: (-2x^2 + 3x - 6) - (-3x - 2) = -2x^2 + 6x - 4. This is correct. The integral of -2x^2 + 6x - 4 from 1 to 2 is indeed 1/3. Wait, let me re-read the problem statement carefully. y = -2x^2 + 3x - 6 and y = -3x - 2. Intersection: -2x^2 + 3x - 6 = -3x - 2 => -2x^2 + 6x - 4 = 0 => x^2 - 3x + 2 = 0 => (x-1)(x-2)=0. Roots are 1 and 2. The parabola opens downward. Between 1 and 2, is the parabola above the line? Test x=1.5: Parabola: -2(2.25) + 4.5 - 6 = -4.5 + 4.5 - 6 = -6. Line: -3(1.5) - 2 = -4.5 - 2 = -6.5. -6 > -6.5, so parabola is on top. The setup is correct. The calculation is correct. Why did I think it was wrong? Let me check the provided solution text again. Step 1 says '-2*(x - 2)*(x - 1), -2*x**2 + 6*x - 4'. This is the integrand. Step 2 says 'They cross at x = 1 and x = 2; between them y = -2*x**2 + 3*x - 6 is on top.' This is correct. Step 3 calculates the integral. The result is 1/3. The solution seems correct. Is there a subtle error? Ah, look at the first equation line: '-2*(x - 2)*(x - 1), -2*x**2 + 6*x - 4'. The comment says 'Set the curves equal: top - bottom, factored.' The expression -2x^2 + 6x - 4 is indeed top - bottom. The factorization is correct. The bounds are correct. The integration is correct. The verdict should be pass? Let me double check the arithmetic. Integral of -2x^2 is -2/3 x^3. Integral of 6x is 3x^2. Integral of -4 is -4x. Evaluated from 1 to 2. At 2: -2/3(8) + 3(4) - 4(2) = -16/3 + 12 - 8 = -16/3 + 4 = -16/3 + 12/3 = -4/3. At 1: -2/3(1) + 3(1) - 4(1) = -2/3 + 3 - 4 = -2/3 - 1 = -5/3. Difference: (-4/3) - (-5/3) = 1/3. The math is correct. The logic is correct. The solution is sound.
  • gpt-oss:20b: pass 2026-09-28
  • qwen3.6:27b-mlx: fail (error) 2026-09-27 — The intersection points are incorrect; solving -2x^2 + 3x - 6 = -3x - 2 yields x = 1 and x = 2, but substituting x=1 gives y=-5 for the parabola and y=-5 for the line, while substituting x=2 gives y=-8 for the parabola and y=-8 for the line. Wait, let me re-check. -2(1)^2 + 3(1) - 6 = -2 + 3 - 6 = -5. -3(1) - 2 = -5. Correct. -2(2)^2 + 3(2) - 6 = -8 + 6 - 6 = -8. -3(2) - 2 = -8. Correct. The intersections are correct. However, the integrand in step 1 is -2x^2 + 6x - 4. This corresponds to (-2x^2 + 3x - 6) - (-3x - 2) = -2x^2 + 6x - 4. This is correct. The integral of -2x^2 + 6x - 4 from 1 to 2 is indeed 1/3. Why did I think it was wrong? Let me re-read the problem statement carefully. y = -2x^2 + 3x - 6 and y = -3x - 2. The setup is correct. The calculation is correct. The answer is correct. Is there a subtle error? "between them y = -2*x**2 + 3*x - 6 is on top". At x=1.5, parabola: -2(2.25) + 4.5 - 6 = -4.5 + 4.5 - 6 = -6. Line: -3(1.5) - 2 = -4.5 - 2 = -6.5. -6 > -6.5, so parabola is on top. This is correct. The solution seems sound. Wait, looking at step 1: "-2*(x - 2)*(x - 1), -2*x**2 + 6*x - 4". The text says "Set the curves equal: top - bottom, factored.". The equation shown is just the factored form and expanded form of the difference. It doesn't explicitly show the subtraction step, but the result is correct. The sentence in step 2 claims they cross at 1 and 2. This is correct. The sentence claims the parabola is on top. This is correct. The integral is correct. The answer is correct. I will mark it as pass.
  • gpt-oss:20b: pass 2026-09-27
  • qwen3.6:27b-mlx: fail (error) 2026-09-26 — The intersection points are incorrect; solving -2x^2 + 3x - 6 = -3x - 2 yields x = 1 and x = 2/3, not x = 1 and x = 2. Additionally, the integrand used (-2x^2 + 6x - 4) corresponds to subtracting y = 2x + 4, not the given line y = -3x - 2.
  • gpt-oss:20b: pass 2026-09-26

Proved: SymPy reduced the difference between the two sides to zero. Checked independently: a separate method, named above, confirmed it. Checked numerically: the two sides agree at every sampled point, which is evidence, not proof. Reviewed: a model or a person read it; that is all a sentence can have. Solution by generator:structured/area_between_curves, checked 2026-09-28 with SymPy 1.14.0.

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