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Fundamental Theorem of Calculus, Part 1

Problem 4.118 · medium

Find \( \displaystyle \dfrac{d}{dx} \displaystyle \int_{2}^{\sqrt{x}} \sin{\left(t^{2} \right)} \, dt \).

By FTC Part 1 and the chain rule, d/dx ∫ from a to u(x) of g(t) dt = g(u(x)) u'(x).
\[ \frac{d}{d x} \sqrt{x} = \frac{1}{2 \sqrt{x}} \]
u'(x).✓ Proved
\[ \frac{\sin{\left(x \right)}}{2 \sqrt{x}} \]
g(u(x)) u'(x).✓ Proved

Answer \( \frac{\sin{\left(x \right)}}{2 \sqrt{x}} \)

Lines: 2 proved, 1 not checked. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation.

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	the integral computed numerically at x = 0.7 ± 10⁻⁶ and differenced agrees

Reviewers

gpt-oss:20b: pass
qwen3.6:27b-mlx: fail (style) — [domain objection, downgraded to style] The solution incorrectly substitutes x for t^2 in the integrand, resulting in sin(x) instead of the correct sin(x). The integrand is sin(t^2), so g(u(x)) should be sin((sqrt(x))^2) = sin(x). Wait, sin((sqrt(x))^2) is indeed sin(x). Let me re-read carefully. g(t) = sin(t^2). u(x) = sqrt(x). g(u(x)) = sin((sqrt(x))^2) = sin(x). The derivative is sin(x)/(2*sqrt(x)). The stated answer is sin(x)/(2*sqrt(x)). The solution says g(u(x))u'(x) is sin(x)/(2*sqrt(x)). This seems correct. Why did I think it was wrong? Let me check the problem statement again. Find d/dx integral from 2 to sqrt(x) of sin(t^2) dt. FTC: d/dx F(u(x)) = f(u(x)) * u'(x). f(t) = sin(t^2). f(u(x)) = sin((sqrt(x))^2) = sin(x). u'(x) = 1/(2sqrt(x)). Result: sin(x)/(2sqrt(x)). The solution is actually correct. Let me look for subtle errors. Line 1 states the formula. Line 2 computes u'. Line 3 computes the product. The algebra is marked proved. The logic is sound. Is there a domain issue? sqrt(x) requires x>=0. The integral lower bound is 2. If x is small, sqrt(x) < 2. The FTC still holds for definite integrals with variable limits as long as the function is continuous. sin(t^2) is continuous everywhere. So the formula holds for all x > 0. The solution is correct.

Every verdict on record (4)

qwen3.6:27b-mlx: fail (style) 2026-09-27 — [domain objection, downgraded to style] The solution incorrectly substitutes x for t^2 in the integrand, resulting in sin(x) instead of the correct sin(x). The integrand is sin(t^2), so g(u(x)) should be sin((sqrt(x))^2) = sin(x). Wait, sin((sqrt(x))^2) is indeed sin(x). Let me re-read carefully. g(t) = sin(t^2). u(x) = sqrt(x). g(u(x)) = sin((sqrt(x))^2) = sin(x). The derivative is sin(x)/(2*sqrt(x)). The stated answer is sin(x)/(2*sqrt(x)). The solution says g(u(x))u'(x) is sin(x)/(2*sqrt(x)). This seems correct. Why did I think it was wrong? Let me check the problem statement again. Find d/dx integral from 2 to sqrt(x) of sin(t^2) dt. FTC: d/dx F(u(x)) = f(u(x)) * u'(x). f(t) = sin(t^2). f(u(x)) = sin((sqrt(x))^2) = sin(x). u'(x) = 1/(2sqrt(x)). Result: sin(x)/(2sqrt(x)). The solution is actually correct. Let me look for subtle errors. Line 1 states the formula. Line 2 computes u'. Line 3 computes the product. The algebra is marked proved. The logic is sound. Is there a domain issue? sqrt(x) requires x>=0. The integral lower bound is 2. If x is small, sqrt(x) < 2. The FTC still holds for definite integrals with variable limits as long as the function is continuous. sin(t^2) is continuous everywhere. So the formula holds for all x > 0. The solution is correct.
gpt-oss:20b: pass 2026-09-27
qwen3.6:27b-mlx: pass 2026-09-27
gpt-oss:20b: pass 2026-09-27

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 2026-09-27 with SymPy 1.14.0.