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Fundamental Theorem of Calculus, Part 1

Problem 4.16 · medium

Find \( \displaystyle \dfrac{d}{dx} \displaystyle \int_{2}^{\sqrt{x}} \sqrt{t^{2} + 1} \, dt \).

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
\[ \frac{d}{d x} \sqrt{x} = \frac{1}{2 \sqrt{x}} \]
u'(x).✓ Proved
\[ \frac{\sqrt{x + 1}}{2 \sqrt{x}} \]
g(u(x)) u'(x).Reviewed

Answer \( \frac{\sqrt{x + 1}}{2 \sqrt{x}} \)

Lines: 1 proved, 2 reviewed. 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	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: pass
qwen3.6:27b-mlx: pass — The solution correctly applies the Fundamental Theorem of Calculus combined with the chain rule. The substitution of u(x) = sqrt(x) into the integrand and the multiplication by the derivative of the upper limit are handled correctly.

Every verdict on record (6)

qwen3.6:27b-mlx: pass 2026-09-28 — The solution correctly applies the Fundamental Theorem of Calculus combined with the chain rule. The substitution of u(x) = sqrt(x) into the integrand and the multiplication by the derivative of the upper limit are handled correctly.
gpt-oss:20b: pass 2026-09-28
qwen3.6:27b-mlx: fail (error) 2026-09-27 — The solution incorrectly identifies g(u(x)) as sqrt(x+1) instead of sqrt(x+1) derived from sqrt(t^2+1) where t=sqrt(x), which is actually sqrt((sqrt(x))^2 + 1) = sqrt(x+1). Wait, sqrt((sqrt(x))^2 + 1) is indeed sqrt(x+1). Let me re-read carefully. g(t) = sqrt(t^2+1). u(x) = sqrt(x). g(u(x)) = sqrt((sqrt(x))^2 + 1) = sqrt(x+1). This is correct. u'(x) = 1/(2*sqrt(x)). The product is sqrt(x+1)/(2*sqrt(x)). The stated answer matches. The sentence in step 1 is correct. Step 2 is correct. Step 3 is 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 sqrt(t^2+1) dt. FTC part 1 says d/dx integral_a^{u(x)} g(t) dt = g(u(x)) u'(x). g(t) = sqrt(t^2+1). u(x) = sqrt(x). g(u(x)) = sqrt(x+1). u'(x) = 1/(2sqrt(x)). Result: sqrt(x+1)/(2sqrt(x)). The solution is correct. I will mark it as pass.
gpt-oss:20b: pass 2026-09-27
qwen3.6:27b-mlx: fail (error) 2026-09-26 — The solution incorrectly evaluates g(u(x)) as sqrt(x+1) instead of the correct sqrt(x+1) derived from substituting u(x)=sqrt(x) into g(t)=sqrt(t^2+1), which yields sqrt((sqrt(x))^2+1) = sqrt(x+1). Wait, sqrt((sqrt(x))^2+1) is indeed sqrt(x+1). Let me re-read carefully. g(t) = sqrt(t^2+1). u(x) = sqrt(x). g(u(x)) = sqrt((sqrt(x))^2 + 1) = sqrt(x+1). The final answer is sqrt(x+1)/(2*sqrt(x)). The stated answer is sqrt(x+1)/(2*sqrt(x)). The solution steps seem correct. Why did I think it was wrong? Let me check the algebra again. g(t) = sqrt(t^2+1). u(x) = sqrt(x). g(u(x)) = sqrt( (sqrt(x))^2 + 1 ) = sqrt(x+1). u'(x) = 1/(2sqrt(x)). Product is sqrt(x+1)/(2sqrt(x)). The solution is correct. I will mark it as pass.
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/ftc_part1, checked 2026-10-13 with SymPy 1.14.0.