Home›Calculus 1›Area between curves›Problem 5.101

Area between curves

Problem 5.101 · medium

Find the area of the region bounded by \( \displaystyle y = - 2 x^{2} - 3 x + 3 \) and \( \displaystyle y = 1 - 3 x \).

\[ \left(2 - 2 x\right) \left(x + 1\right) = 2 - 2 x^{2} \]
Set the curves equal: top − bottom, factored.✓ Proved
They cross at x = -1 and x = 1; between them y = -2*x**2 - 3*x + 3 is on top.
\[ \int\limits_{-1}^{1} \left(2 - 2 x^{2}\right)\, dx = \frac{8}{3} \]
Area = ∫ (top − bottom) dx.✓ Proved

Answer \( \frac{8}{3} \)

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.

The full receipt

Line	Status	Checked by	Detail
1	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
2	Not checked	—	a sentence; read, not computed
3	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
answer, a second way	✓ Checked independently	sympy 1.14.0 + mpmath 1.3.0	numerical quadrature of \|top − bottom\| gives the same area

Reviewers

gpt-oss:20b: pass
qwen3.6:27b-mlx: fail (error) — The intersection points are calculated incorrectly; solving -2x^2 - 3x + 3 = 1 - 3x yields x = ±1, but substituting x=1 gives y=1-3=-2 for the line and y=-2-3+3=-2 for the parabola, which is correct, wait. Let's re-solve: -2x^2 - 3x + 3 = 1 - 3x => -2x^2 + 2 = 0 => x^2 = 1 => x = ±1. The intersection points are actually correct. However, the integrand in step 1 is wrong. Top - Bottom = (-2x^2 - 3x + 3) - (1 - 3x) = -2x^2 + 2. The solution uses 2 - 2x^2, which is correct. Wait, let me re-read step 1. It says '-2*(x - 1)*(x + 1), 2 - 2*x**2'. This is correct. Let me check the intersection again. -2x^2 - 3x + 3 = 1 - 3x. Add 3x to both sides: -2x^2 + 3 = 1. Subtract 1: -2x^2 + 2 = 0. 2x^2 = 2. x^2 = 1. x = 1, -1. The intersections are correct. The integrand is correct. The integral of 2 - 2x^2 from -1 to 1 is [2x - 2/3 x^3] from -1 to 1 = (2 - 2/3) - (-2 + 2/3) = 4/3 - (-4/3) = 8/3. The solution is actually correct. Why did I think it was wrong? Let me look closer at the problem statement. y = -2x^2 - 3x + 3 and y = 1 - 3x. The solution says 'They cross at x = -1 and x = 1'. This is correct. 'Between them y = -2*x**2 - 3*x + 3 is on top'. Let's check x=0. Parabola: 3. Line: 1. 3 > 1. Correct. The integral is correct. The answer is correct. Is there a subtle error? Ah, look at step 1 text: '-2*(x - 1)*(x + 1), 2 - 2*x**2'. This is just showing the factored form and expanded form. It seems fine. Wait, is the prompt tricking me? Let me re-read the prompt carefully. 'Find the area of the region bounded by...'. The solution seems perfectly valid. Let me check if I missed a sign. -2x^2 - 3x + 3 - (1 - 3x) = -2x^2 + 2. Yes. Integral from -1 to 1 of (2 - 2x^2) dx. Yes. Result 8/3. Yes. Okay, maybe I should look for a 'style' or 'misleading' issue? No, it's concise and correct. Wait, did I misread the problem? y = -2x^2 - 3x + 3. y = 1 - 3x. Okay. What if the curves don't bound a single region? They intersect at two points, forming a single enclosed region. Okay. Is it possible the user made a typo in the problem statement provided to me vs the solution? No, I must review the solution against the problem. The solution is correct. I will mark it as pass.

Every verdict on record (4)

qwen3.6:27b-mlx: fail (error) 2026-09-26 — The intersection points are calculated incorrectly; solving -2x^2 - 3x + 3 = 1 - 3x yields x = ±1, but substituting x=1 gives y=1-3=-2 for the line and y=-2-3+3=-2 for the parabola, which is correct, wait. Let's re-solve: -2x^2 - 3x + 3 = 1 - 3x => -2x^2 + 2 = 0 => x^2 = 1 => x = ±1. The intersection points are actually correct. However, the integrand in step 1 is wrong. Top - Bottom = (-2x^2 - 3x + 3) - (1 - 3x) = -2x^2 + 2. The solution uses 2 - 2x^2, which is correct. Wait, let me re-read step 1. It says '-2*(x - 1)*(x + 1), 2 - 2*x**2'. This is correct. Let me check the intersection again. -2x^2 - 3x + 3 = 1 - 3x. Add 3x to both sides: -2x^2 + 3 = 1. Subtract 1: -2x^2 + 2 = 0. 2x^2 = 2. x^2 = 1. x = 1, -1. The intersections are correct. The integrand is correct. The integral of 2 - 2x^2 from -1 to 1 is [2x - 2/3 x^3] from -1 to 1 = (2 - 2/3) - (-2 + 2/3) = 4/3 - (-4/3) = 8/3. The solution is actually correct. Why did I think it was wrong? Let me look closer at the problem statement. y = -2x^2 - 3x + 3 and y = 1 - 3x. The solution says 'They cross at x = -1 and x = 1'. This is correct. 'Between them y = -2*x**2 - 3*x + 3 is on top'. Let's check x=0. Parabola: 3. Line: 1. 3 > 1. Correct. The integral is correct. The answer is correct. Is there a subtle error? Ah, look at step 1 text: '-2*(x - 1)*(x + 1), 2 - 2*x**2'. This is just showing the factored form and expanded form. It seems fine. Wait, is the prompt tricking me? Let me re-read the prompt carefully. 'Find the area of the region bounded by...'. The solution seems perfectly valid. Let me check if I missed a sign. -2x^2 - 3x + 3 - (1 - 3x) = -2x^2 + 2. Yes. Integral from -1 to 1 of (2 - 2x^2) dx. Yes. Result 8/3. Yes. Okay, maybe I should look for a 'style' or 'misleading' issue? No, it's concise and correct. Wait, did I misread the problem? y = -2x^2 - 3x + 3. y = 1 - 3x. Okay. What if the curves don't bound a single region? They intersect at two points, forming a single enclosed region. Okay. Is it possible the user made a typo in the problem statement provided to me vs the solution? No, I must review the solution against the problem. The solution is correct. I will mark it as pass.
gpt-oss:20b: pass 2026-09-26
qwen3.6:27b-mlx: fail (error) 2026-09-26 — The setup is incorrect because the intersection points are not x = -1 and x = 1; solving -2x^2 - 3x + 3 = 1 - 3x yields x = ±1/sqrt(2). Consequently, the integrand and bounds are wrong.
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 2026-09-26 with SymPy 1.14.0.