Accumulation functions
Problem 4.1082 · easy
Use the Fundamental Theorem of Calculus to write \( \displaystyle F(x) = \displaystyle \int_{1}^{x} \sqrt{t} \, dt \) as a function of \( \displaystyle x \).
- \[ \frac{d}{d t} \frac{2 t^{\frac{3}{2}}}{3} = \sqrt{t} \]An antiderivative of the integrand is 2*t**(3/2)/3.✓ Proved
- \[ - \left. \frac{2 t^{\frac{3}{2}}}{3} \right|_{\substack{ t=1 }} + \left. \frac{2 t^{\frac{3}{2}}}{3} \right|_{\substack{ t=x }} = \frac{2 x^{\frac{3}{2}}}{3} - \frac{2}{3} \]F(x) = G(x) − G(a).✓ Proved
- \[ \frac{d}{d x} \left(\frac{2 x^{\frac{3}{2}}}{3} - \frac{2}{3}\right) = \sqrt{x} \]F′(x) = f(x), as FTC Part 1 says.✓ Proved
- \[ \left. \frac{2 x^{\frac{3}{2}}}{3} - \frac{2}{3} \right|_{\substack{ x=1 }} = 0 \]And F(1) = 0.✓ Proved
Answer \( F(x) = \frac{2 x^{\frac{3}{2}}}{3} - \frac{2}{3} \)
✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. 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 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 2 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 4 | ✓ 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 from a to three values of x |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: pass
Every verdict on record (6)
qwen3.6:27b-mlx: pass 2026-10-17gpt-oss:20b: pass 2026-10-17qwen3.6:27b-mlx: pass 2026-10-14 — The solution correctly identifies an antiderivative and applies the Fundamental Theorem of Calculus Part 2 to evaluate the definite integral. The verification steps are consistent with the result.gpt-oss:20b: pass 2026-10-14qwen3.6:27b-mlx: pass 2026-10-14 — The solution correctly identifies an antiderivative and applies the Fundamental Theorem of Calculus Part 2 to evaluate the definite integral. The verification steps in lines 3 and 4 are consistent with the problem statement.gpt-oss:20b: pass 2026-10-14
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_accumulation, checked 2026-10-17 with SymPy 1.14.0.