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Fundamental Theorem of Calculus, Part 1

Problem 4.118 · medium

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

Lines: 1 proved, 2 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
LineStatusChecked byDetail
1Not checked—a sentence; read, not computed
2✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
3Not checked—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 independentlysympy 1.14.0 + mpmath 1.3.0the integral computed numerically at x = 0.7 ± 10⁻⁶ and differenced agrees

Reviewers

  • gpt-oss:20b: pass
  • qwen3.6:27b-mlx: fail (error) — The solution incorrectly substitutes x for u(x)^2 in the integrand. Since u(x) = sqrt(x), g(u(x)) should be sin((sqrt(x))^2) = sin(x), which happens to be correct, but the logic implies g(t) = sin(t) rather than sin(t^2). Wait, if g(t) = sin(t^2), then g(u(x)) = sin((sqrt(x))^2) = sin(x). The result is actually correct. Let me re-read carefully. The problem is d/dx integral from 2 to sqrt(x) of sin(t^2) dt. FTC says derivative is g(u(x)) * u'(x). g(t) = sin(t^2). u(x) = sqrt(x). g(u(x)) = sin((sqrt(x))^2) = sin(x). u'(x) = 1/(2*sqrt(x)). Result: sin(x)/(2*sqrt(x)). The stated answer is sin(x)/(2*sqrt(x)). The solution steps: 1. States formula. 2. Computes u'(x). 3. States the final result. Is there an error? Step 3 says: `sin(x)/(2*sqrt(x)), sin(x)/(2*sqrt(x)) -- g(u(x)) u'(x)`. This implies g(u(x)) * u'(x) equals sin(x)/(2*sqrt(x)). Since g(u(x)) = sin(x) and u'(x) = 1/(2*sqrt(x)), the product is indeed sin(x)/(2*sqrt(x)). Wait, did the student make a mistake in reasoning that just happened to be right? Or is the notation misleading? The sentence in step 1 is correct. The equation in step 2 is correct. The equation in step 3 is correct. Let's look closer at Step 3. It labels the result as `g(u(x)) u'(x)`. Does `g(u(x))` equal `sin(x)`? Yes, because `u(x) = sqrt(x)` and `g(t) = sin(t^2)`, so `g(sqrt(x)) = sin((sqrt(x))^2) = sin(x)`. Is there any subtlety? If the student thought `g(t) = sin(t)`, then `g(u(x)) = sin(sqrt(x))`, and the answer would be `sin(sqrt(x))/(2*sqrt(x))`. This is wrong. But the answer provided is `sin(x)/(2*sqrt(x))`. This is the correct answer. So the solution is correct. Why would it fail? Maybe the prompt implies I should find what is wrong. If nothing is wrong, I pass. Let's double check the FTC application. F(x) = integral_2^{sqrt(x)} sin(t^2) dt. Let G be antiderivative of sin(t^2). F(x) = G(sqrt(x)) - G(2). F'(x) = G'(sqrt(x)) * (sqrt(x))' = sin((sqrt(x))^2) * (1/(2*sqrt(x))) = sin(x)/(2*sqrt(x)). The solution is sound. The intermediate step of squaring the sqrt is implicit in the jump from the formula to the final answer, but the final answer is correct. The label `g(u(x)) u'(x)` is correct for the expression shown. Verdict: pass.
Every verdict on record (6)
  • qwen3.6:27b-mlx: fail (error) 2026-09-28 — The solution incorrectly substitutes x for u(x)^2 in the integrand. Since u(x) = sqrt(x), g(u(x)) should be sin((sqrt(x))^2) = sin(x), which happens to be correct, but the logic implies g(t) = sin(t) rather than sin(t^2). Wait, if g(t) = sin(t^2), then g(u(x)) = sin((sqrt(x))^2) = sin(x). The result is actually correct. Let me re-read carefully. The problem is d/dx integral from 2 to sqrt(x) of sin(t^2) dt. FTC says derivative is g(u(x)) * u'(x). g(t) = sin(t^2). u(x) = sqrt(x). g(u(x)) = sin((sqrt(x))^2) = sin(x). u'(x) = 1/(2*sqrt(x)). Result: sin(x)/(2*sqrt(x)). The stated answer is sin(x)/(2*sqrt(x)). The solution steps: 1. States formula. 2. Computes u'(x). 3. States the final result. Is there an error? Step 3 says: `sin(x)/(2*sqrt(x)), sin(x)/(2*sqrt(x)) -- g(u(x)) u'(x)`. This implies g(u(x)) * u'(x) equals sin(x)/(2*sqrt(x)). Since g(u(x)) = sin(x) and u'(x) = 1/(2*sqrt(x)), the product is indeed sin(x)/(2*sqrt(x)). Wait, did the student make a mistake in reasoning that just happened to be right? Or is the notation misleading? The sentence in step 1 is correct. The equation in step 2 is correct. The equation in step 3 is correct. Let's look closer at Step 3. It labels the result as `g(u(x)) u'(x)`. Does `g(u(x))` equal `sin(x)`? Yes, because `u(x) = sqrt(x)` and `g(t) = sin(t^2)`, so `g(sqrt(x)) = sin((sqrt(x))^2) = sin(x)`. Is there any subtlety? If the student thought `g(t) = sin(t)`, then `g(u(x)) = sin(sqrt(x))`, and the answer would be `sin(sqrt(x))/(2*sqrt(x))`. This is wrong. But the answer provided is `sin(x)/(2*sqrt(x))`. This is the correct answer. So the solution is correct. Why would it fail? Maybe the prompt implies I should find what is wrong. If nothing is wrong, I pass. Let's double check the FTC application. F(x) = integral_2^{sqrt(x)} sin(t^2) dt. Let G be antiderivative of sin(t^2). F(x) = G(sqrt(x)) - G(2). F'(x) = G'(sqrt(x)) * (sqrt(x))' = sin((sqrt(x))^2) * (1/(2*sqrt(x))) = sin(x)/(2*sqrt(x)). The solution is sound. The intermediate step of squaring the sqrt is implicit in the jump from the formula to the final answer, but the final answer is correct. The label `g(u(x)) u'(x)` is correct for the expression shown. Verdict: pass.
  • gpt-oss:20b: pass 2026-09-28
  • 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 2026-10-13 with SymPy 1.14.0.

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