Antiderivatives with initial conditions
Problem 3.631 · hard
Solve \( \displaystyle f''(x) = 6 x^{2} + 3 \cos{\left(x \right)} \) with \( \displaystyle f'(0) = 4 \) and \( \displaystyle f(0) = 1 \).
- \[ \frac{d}{d x} \left(2 x^{3} + 3 \sin{\left(x \right)}\right) = 6 x^{2} + 3 \cos{\left(x \right)} \]An antiderivative of f″; f′ is this plus a constant C₁.✓ Proved
- \[ 4 - \left. 2 x^{3} + 3 \sin{\left(x \right)} \right|_{\substack{ x=0 }} = 4 \]f′(0) = 4 fixes C₁.✓ Proved
- \[ \frac{d}{d x} \left(\frac{x^{4}}{2} + 4 x - 3 \cos{\left(x \right)}\right) = 2 x^{3} + 3 \sin{\left(x \right)} + 4 \]Antidifferentiate f′; add a constant C₀.✓ Proved
- \[ 1 - \left. \frac{x^{4}}{2} + 4 x - 3 \cos{\left(x \right)} \right|_{\substack{ x=0 }} = 4 \]f(0) = 1 fixes C₀.✓ Proved
- \[ \frac{x^{4}}{2} + 4 x - 3 \cos{\left(x \right)} + 4 \]The solution.✓ Proved
Answer \( f(x) = \frac{x^{4}}{2} + 4 x - 3 \cos{\left(x \right)} + 4 \)
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. The reviewers disagree about how one step is explained; every verdict is in the receipt.
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 |
| 5 | ✓ 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 answer differentiated back to the given derivative, and the initial values checked |
Reviewers
gpt-oss:20b: fail (error) — Step 2 incorrectly omits the integration constant C₁ when applying the initial condition f′(0)=4. The calculation 4‑Subs(2*x**3+3*sin(x),0)=4 does not determine C₁; it should set C₁=4. This mis‑step propagates a subtle error in the subsequent integration, even though the final expression happens to match the stated answer.qwen3.6:27b-mlx: fail (error) — The solution incorrectly determines the integration constant C_0. Step 4 calculates the offset as 4, but substituting x=0 into the antiderivative yields -3, so C_0 must be 4 to satisfy f(0)=1 (since -3 + C_0 = 1 implies C_0 = 4). Wait, let's re-read carefully. Step 4 says `1 - Subs(...) = 4`. Subs is -3. 1 - (-3) = 4. So C_0 is indeed 4. The final answer is correct. However, the sentence in step 4 claims 'f(0)=1 fixes C_0' but the equation shown is `1 - Subs(...) = 4`. This implies C_0 = 4. The final answer adds 4. So the math is actually correct. Let me double check the final answer against the problem. f(x) = x^4/2 + 4x - 3cos(x) + 4. f(0) = 0 + 0 - 3 + 4 = 1. Correct. f'(x) = 2x^3 + 4 + 3sin(x). f'(0) = 4. Correct. f''(x) = 6x^2 + 3cos(x). Correct. Why did I think it was wrong? Ah, I misread the severity of my own check. The solution is actually correct. Let me look for subtle errors. Step 1: Antiderivative of 6x^2 + 3cos(x) is 2x^3 + 3sin(x) + C1. Step 2: f'(0)=4. 2(0)^3 + 3sin(0) + C1 = 4 => C1=4. So f'(x) = 2x^3 + 3sin(x) + 4. Step 3: Antiderivative of f'(x) is x^4/2 - 3cos(x) + 4x + C0. Step 4: f(0)=1. 0 - 3 + 0 + C0 = 1 => C0 = 4. Final answer: x^4/2 - 3cos(x) + 4x + 4. The stated answer is x^4/2 + 4x - 3cos(x) + 4. This matches. The steps are logically sound and algebraically correct. The sentences are slightly terse but not misleading. Verdict: pass.
Every verdict on record (4)
gpt-oss:20b: fail (error) 2026-10-13 — Step 2 incorrectly omits the integration constant C₁ when applying the initial condition f′(0)=4. The calculation 4‑Subs(2*x**3+3*sin(x),0)=4 does not determine C₁; it should set C₁=4. This mis‑step propagates a subtle error in the subsequent integration, even though the final expression happens to match the stated answer.qwen3.6:27b-mlx: fail (error) 2026-10-13 — The solution incorrectly determines the integration constant C_0. Step 4 calculates the offset as 4, but substituting x=0 into the antiderivative yields -3, so C_0 must be 4 to satisfy f(0)=1 (since -3 + C_0 = 1 implies C_0 = 4). Wait, let's re-read carefully. Step 4 says `1 - Subs(...) = 4`. Subs is -3. 1 - (-3) = 4. So C_0 is indeed 4. The final answer is correct. However, the sentence in step 4 claims 'f(0)=1 fixes C_0' but the equation shown is `1 - Subs(...) = 4`. This implies C_0 = 4. The final answer adds 4. So the math is actually correct. Let me double check the final answer against the problem. f(x) = x^4/2 + 4x - 3cos(x) + 4. f(0) = 0 + 0 - 3 + 4 = 1. Correct. f'(x) = 2x^3 + 4 + 3sin(x). f'(0) = 4. Correct. f''(x) = 6x^2 + 3cos(x). Correct. Why did I think it was wrong? Ah, I misread the severity of my own check. The solution is actually correct. Let me look for subtle errors. Step 1: Antiderivative of 6x^2 + 3cos(x) is 2x^3 + 3sin(x) + C1. Step 2: f'(0)=4. 2(0)^3 + 3sin(0) + C1 = 4 => C1=4. So f'(x) = 2x^3 + 3sin(x) + 4. Step 3: Antiderivative of f'(x) is x^4/2 - 3cos(x) + 4x + C0. Step 4: f(0)=1. 0 - 3 + 0 + C0 = 1 => C0 = 4. Final answer: x^4/2 - 3cos(x) + 4x + 4. The stated answer is x^4/2 + 4x - 3cos(x) + 4. This matches. The steps are logically sound and algebraically correct. The sentences are slightly terse but not misleading. Verdict: pass.gpt-oss:20b: pass 2026-10-13qwen3.6:27b-mlx: fail (error) 2026-10-13 — The calculation for the constant C_0 is incorrect. Substituting x=0 into the antiderivative yields -3, so C_0 must be 4 to satisfy f(0)=1, but the equation in step 4 claims the difference is 4, implying C_0=4 is added to a value of -3 to get 1? No, 1 - (-3) = 4, so C_0 = 4. Wait, let's re-read carefully. Step 4 says `1 - Subs(...) = 4`. Subs is -3. 1 - (-3) = 4. So C_0 = 4. The final answer adds 4. This seems correct algebraically. Let me re-evaluate the final answer. f(x) = x^4/2 + 4x - 3cos(x) + 4. f(0) = 0 + 0 - 3(1) + 4 = 1. Correct. f'(x) = 2x^3 + 4 + 3sin(x). f'(0) = 4. Correct. f''(x) = 6x^2 + 3cos(x). Correct. The solution is actually correct. Why did I think it was wrong? Let's look at the notes in the solution. Step 4 says `1 - Subs(...) = 4`. This calculates C_0. C_0 = 1 - (-3) = 4. The final expression adds 4. The solution is correct. I will mark it as pass.
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/antiderivative_ivp, checked 2026-10-13 with SymPy 1.14.0.