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Definite integrals from area formulas

Problem 4.1184 · easy Video

Evaluate \( \displaystyle \int_{0}^{4} \sqrt{16 - x^{2}} \, dx \) using a formula from geometry.

The region between the graph and the x-axis is a quarter circle of radius 4.
\[ \frac{4^{2} \pi}{4} = 4 \pi \]
Its (signed) area.✓ Proved
\[ \int\limits_{0}^{4} \sqrt{16 - x^{2}}\, dx = 4 \pi \]
Integrating directly gives the same.✓ Proved

Answer \( 4 \pi \)

The graph of f(x) = sqrt(16 - x²) from x = 0 to x = 4. Rectangles under the curve get thinner until they fill the area, 4pi.

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: inconclusive — reviewer returned a non-object

Every verdict on record (6)

qwen3.6:27b-mlx: inconclusive 2026-10-18 — reviewer returned a non-object
gpt-oss:20b: pass 2026-10-18
qwen3.6:27b-mlx: inconclusive 2026-10-16 — reviewer returned a non-object
gpt-oss:20b: pass 2026-10-16
qwen3.6:27b-mlx: inconclusive 2026-10-15 — reviewer returned a non-object
gpt-oss:20b: pass 2026-10-15

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 2026-10-18 with SymPy 1.14.0.