Home›Calculus 1›Definite integrals from area formulas›Problem 4.1376

Definite integrals from area formulas

Problem 4.1376 · easy Video

Evaluate \( \displaystyle \int_{0}^{2} - x \, dx \) using a formula from geometry.

The region between the graph and the x-axis is a trapezoid (a triangle, since one side has height 0), below the axis.
\[ \frac{\left(-2 + 0\right) \left(0 + 2\right)}{2} = -2 \]
Its (signed) area.✓ Proved
\[ \int\limits_{0}^{2} \left(- x\right)\, dx = -2 \]
Integrating directly gives the same.✓ Proved

Answer \( -2 \)

The graph of f(x) = -x from x = 0 to x = 2. Rectangles under the curve get thinner until they fill the area, -2. Area below the x-axis counts as negative.

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

Line	Status	Checked by	Detail
1	Not checked	—	a sentence; read, not computed
2	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
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

Reviewers

gpt-oss:20b: pass
qwen3.6:27b-mlx: fail (style) — [domain objection, downgraded to style] The solution incorrectly identifies the region as a trapezoid. The region bounded by y = -x, the x-axis, and x = 0 to 2 is a triangle, not a general trapezoid (though a triangle is a degenerate trapezoid, the specific formula used is for a trapezoid, which is unnecessarily complex and potentially confusing when a simple triangle area formula suffices). More critically, the formula applied in step 2 is the area of a trapezoid with parallel vertical sides? No, the formula (b1+b2)*h/2 is for a trapezoid with parallel sides b1, b2 and height h. Here, the 'parallel sides' are the vertical segments at x=0 (length 0) and x=2 (length 2), and the 'height' is the horizontal distance 2. This is technically correct for the area of the triangle, but describing it as a trapezoid is misleading terminology for a student learning geometry formulas. However, the bigger issue is that the problem asks to use a formula from geometry. The standard geometric formula for the area of a triangle is 1/2 * base * height. The solution uses the trapezoid formula. While mathematically equivalent, it's a poor pedagogical choice. But is it an error? A triangle is a trapezoid. The calculation is correct. Let's look closer. Is there a more fundamental error? The integral is negative. The area is positive. The solution says 'Its (signed) area'. The geometric formula gives positive area. The solution assigns the negative sign manually. This is acceptable. However, calling the region a 'trapezoid' is technically incorrect in standard elementary geometry contexts where a trapezoid is defined as a quadrilateral with at least one pair of parallel sides, but usually implies 4 distinct vertices or non-degenerate parallel sides. A triangle has 3 vertices. While some definitions allow triangles as degenerate trapezoids, it is misleading to a student. Let's check the formula application again. (0 + 2) * 2 / 2 = 2. The signed area is -2. The equation shows (-2 + 0)... wait. The heights are y(0)=0 and y(2)=-2. The lengths of the vertical segments are 0 and 2. The formula for area of a trapezoid with vertical parallel sides of length a and b and width w is (a+b)w/2. Here a=0, b=2, w=2. Area = (0+2)*2/2 = 2. The signed area is -2. The equation in step 2 is `(-2 + 0)*(0 + 2)/2 = -2`. This mixes signed values into the geometric formula. The geometric formula for area uses lengths (positive). Using signed values in the trapezoid area formula is not standard 'geometry' but rather algebraic manipulation disguised as geometry. A strict 'formula from geometry' would be Area = 1/2 * base * height = 1/2 * 2 * 2 = 2, then apply sign. The solution's equation `(-2 + 0)...` suggests using the y-values directly in the trapezoid formula. This is valid for the integral of a linear function, but it's not a 'formula from geometry' in the standard sense (which deals with positive lengths). It's an algebraic coincidence that the trapezoid rule works for linear functions. The prompt asks to use a formula from geometry. The standard geometric formula is for triangles or rectangles. Using the trapezoid rule (which is a numerical integration method or derived from geometry) is borderline. But the main error is likely the classification. A triangle is not typically called a trapezoid in introductory calculus/geometry contexts. It's a triangle. Using the trapezoid formula is overcomplicating and terminologically loose. I will mark it as misleading/error because it teaches a student to call a triangle a trapezoid and to plug signed values into geometric area formulas.

Every verdict on record (6)

qwen3.6:27b-mlx: fail (style) 2026-10-29 — [domain objection, downgraded to style] The solution incorrectly identifies the region as a trapezoid. The region bounded by y = -x, the x-axis, and x = 0 to 2 is a triangle, not a general trapezoid (though a triangle is a degenerate trapezoid, the specific formula used is for a trapezoid, which is unnecessarily complex and potentially confusing when a simple triangle area formula suffices). More critically, the formula applied in step 2 is the area of a trapezoid with parallel vertical sides? No, the formula (b1+b2)*h/2 is for a trapezoid with parallel sides b1, b2 and height h. Here, the 'parallel sides' are the vertical segments at x=0 (length 0) and x=2 (length 2), and the 'height' is the horizontal distance 2. This is technically correct for the area of the triangle, but describing it as a trapezoid is misleading terminology for a student learning geometry formulas. However, the bigger issue is that the problem asks to use a formula from geometry. The standard geometric formula for the area of a triangle is 1/2 * base * height. The solution uses the trapezoid formula. While mathematically equivalent, it's a poor pedagogical choice. But is it an error? A triangle is a trapezoid. The calculation is correct. Let's look closer. Is there a more fundamental error? The integral is negative. The area is positive. The solution says 'Its (signed) area'. The geometric formula gives positive area. The solution assigns the negative sign manually. This is acceptable. However, calling the region a 'trapezoid' is technically incorrect in standard elementary geometry contexts where a trapezoid is defined as a quadrilateral with at least one pair of parallel sides, but usually implies 4 distinct vertices or non-degenerate parallel sides. A triangle has 3 vertices. While some definitions allow triangles as degenerate trapezoids, it is misleading to a student. Let's check the formula application again. (0 + 2) * 2 / 2 = 2. The signed area is -2. The equation shows (-2 + 0)... wait. The heights are y(0)=0 and y(2)=-2. The lengths of the vertical segments are 0 and 2. The formula for area of a trapezoid with vertical parallel sides of length a and b and width w is (a+b)w/2. Here a=0, b=2, w=2. Area = (0+2)*2/2 = 2. The signed area is -2. The equation in step 2 is `(-2 + 0)*(0 + 2)/2 = -2`. This mixes signed values into the geometric formula. The geometric formula for area uses lengths (positive). Using signed values in the trapezoid area formula is not standard 'geometry' but rather algebraic manipulation disguised as geometry. A strict 'formula from geometry' would be Area = 1/2 * base * height = 1/2 * 2 * 2 = 2, then apply sign. The solution's equation `(-2 + 0)...` suggests using the y-values directly in the trapezoid formula. This is valid for the integral of a linear function, but it's not a 'formula from geometry' in the standard sense (which deals with positive lengths). It's an algebraic coincidence that the trapezoid rule works for linear functions. The prompt asks to use a formula from geometry. The standard geometric formula is for triangles or rectangles. Using the trapezoid rule (which is a numerical integration method or derived from geometry) is borderline. But the main error is likely the classification. A triangle is not typically called a trapezoid in introductory calculus/geometry contexts. It's a triangle. Using the trapezoid formula is overcomplicating and terminologically loose. I will mark it as misleading/error because it teaches a student to call a triangle a trapezoid and to plug signed values into geometric area formulas.
gpt-oss:20b: pass 2026-10-29
qwen3.6:27b-mlx: inconclusive 2026-10-28 — reviewer returned a non-object
gpt-oss:20b: pass 2026-10-28
qwen3.6:27b-mlx: fail (error) 2026-10-28 — The solution incorrectly identifies the region as a trapezoid; the region bounded by y=-x, the x-axis, and x=0 to x=2 is a triangle. Furthermore, the formula used in step 2 is the area of a trapezoid with parallel vertical sides, but the calculation (-2+0)*(2)/2 implies using the x-interval as the height and y-values as bases, which is dimensionally inconsistent with the standard trapezoid area formula (average of parallel sides times height) unless interpreted loosely as 'average height times width', but the geometric description is flawed.
gpt-oss:20b: pass 2026-10-28

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/integral_geometry, checked 2026-10-29 with SymPy 1.14.0.