Fundamental Theorem of Calculus, Part 1
Problem 4.30 · medium
Find \( \displaystyle \dfrac{d}{dx} \displaystyle \int_{0}^{\sin{\left(x \right)}} e^{- t^{2}} \, dt \).
- By FTC Part 1 and the chain rule, d/dx ∫ from a to u(x) of g(t) dt = g(u(x)) u'(x).Reviewed
- \[ \frac{d}{d x} \sin{\left(x \right)} = \cos{\left(x \right)} \]u'(x).✓ Proved
- \[ e^{- \sin^{2}{\left(x \right)}} \cos{\left(x \right)} \]g(u(x)) u'(x).Reviewed
Answer \( e^{- \sin^{2}{\left(x \right)}} \cos{\left(x \right)} \)
✓ Nihil obstat Lines: 1 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. How we verify
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | Reviewed | reviewers: gpt-oss:20b, qwen3.6:27b-mlx | a sentence; read, not computed |
| 2 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 3 | Reviewed | reviewers: gpt-oss:20b, qwen3.6:27b-mlx | both sides are the same expression: the line states a value, and the working that produced it is not on the line |
| answer, a second way | ✓ Checked independently | sympy 1.14.0 + mpmath 1.3.0 | the integral computed numerically at x = 0.7 ± 10⁻⁶ and differenced agrees |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: pass
Every verdict on record (6)
qwen3.6:27b-mlx: pass 2026-09-28gpt-oss:20b: pass 2026-09-28qwen3.6:27b-mlx: pass 2026-09-27 — The solution correctly applies the Fundamental Theorem of Calculus Part 1 combined with the Chain Rule. The setup is accurate, and the algebraic steps are verified.gpt-oss:20b: pass 2026-09-27qwen3.6:27b-mlx: pass 2026-09-26gpt-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/ftc_part1, checked 2026-10-13 with SymPy 1.14.0.