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).
- \[ \frac{d}{d x} \sqrt{x} = \frac{1}{2 \sqrt{x}} \]u'(x).✓ Proved
- \[ \frac{\sqrt{x + 1}}{2 \sqrt{x}} \]g(u(x)) u'(x).✓ Proved
Answer \( \frac{\sqrt{x + 1}}{2 \sqrt{x}} \)
Lines: 2 proved, 1 not checked. The answer was also checked a second way, without looking at the solution. The explanation has not been reviewed yet.
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 |
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 2026-09-26 with SymPy 1.14.0.