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Question 1318: Zero-Constant Antiderivative of a Monomial

On this page
  1. Original question and givens
  2. Definition and scope
  3. Step-by-step derivation
  4. Reverse checks
  5. Result
  6. This question review and source binding

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← Question derivation index · Original question batch

R0001-01318 · q-20261007-advance-20261008-r06-antiderivative-0018

Original question and givens

On all R, f(x)=(-18)x^5, with 0≤n≤5 and x^0=1 even at zero. Give ascending reduced rational coefficients of its zero-constant primitive.

  • G001 — On all R, f(x)=(-18)x^5, with 0≤n≤5 and x^0=1 even at zero. Give ascending reduced rational coefficients of its zero-constant primitive.

Definition and scope

Use the original question domain, assumptions and cited MKO definitions. This item has second-context mathematical and bilingual review, with the model, answer and every arithmetic step independently recomputed from the original parameters; the source question remains a candidate.

Step-by-step derivation

S001 · implies

Before

domain=R; coefficient lists are in ascending powers

After

constant term first; rational pairs=[numerator,denominator]

Conditions: none

Fix coefficient order;do not read ascending coefficients as descending or swap numerator and denominator.

Step identifiers and dependencies

Rule: polynomial_representation · Uses: G001

S002 · implies

Before

f coefficients ascending=[0, 0, 0, 0, 0, -18]

After

seek F₀ with constant coefficient0 and F₀′=f

Conditions: none

Antiderivatives are not generally unique; this task additionally fixes the constant coefficient at0.

Step identifiers and dependencies

Rule: canonical_primitive_domain · Uses: G001, S001

S003 · calculation

Before

primitive coefficient at degree1 := div(0, 1)

After

primitive coefficient at degree1 = 0

Conditions: none

Since (x^(k+1))′=(k+1)x^k,divide the coefficient by the positive nonzero integer k+1.

Step identifiers and dependencies

Rule: exact_rational_arithmetic · Uses: G001, S002

S004 · calculation

Before

differentiate back coefficient0 := mul(0, 1)

After

differentiate back coefficient0 = 0

Conditions: none

Differentiation multiplies back by k+1 and must recover the original coefficient.

Step identifiers and dependencies

Rule: exact_rational_arithmetic · Uses: G001, S003

S005 · calculation

Before

primitive coefficient at degree2 := div(0, 2)

After

primitive coefficient at degree2 = 0

Conditions: none

Since (x^(k+1))′=(k+1)x^k,divide the coefficient by the positive nonzero integer k+1.

Step identifiers and dependencies

Rule: exact_rational_arithmetic · Uses: G001, S004

S006 · calculation

Before

differentiate back coefficient1 := mul(0, 2)

After

differentiate back coefficient1 = 0

Conditions: none

Differentiation multiplies back by k+1 and must recover the original coefficient.

Step identifiers and dependencies

Rule: exact_rational_arithmetic · Uses: G001, S005

S007 · calculation

Before

primitive coefficient at degree3 := div(0, 3)

After

primitive coefficient at degree3 = 0

Conditions: none

Since (x^(k+1))′=(k+1)x^k,divide the coefficient by the positive nonzero integer k+1.

Step identifiers and dependencies

Rule: exact_rational_arithmetic · Uses: G001, S006

S008 · calculation

Before

differentiate back coefficient2 := mul(0, 3)

After

differentiate back coefficient2 = 0

Conditions: none

Differentiation multiplies back by k+1 and must recover the original coefficient.

Step identifiers and dependencies

Rule: exact_rational_arithmetic · Uses: G001, S007

S009 · calculation

Before

primitive coefficient at degree4 := div(0, 4)

After

primitive coefficient at degree4 = 0

Conditions: none

Since (x^(k+1))′=(k+1)x^k,divide the coefficient by the positive nonzero integer k+1.

Step identifiers and dependencies

Rule: exact_rational_arithmetic · Uses: G001, S008

S010 · calculation

Before

differentiate back coefficient3 := mul(0, 4)

After

differentiate back coefficient3 = 0

Conditions: none

Differentiation multiplies back by k+1 and must recover the original coefficient.

Step identifiers and dependencies

Rule: exact_rational_arithmetic · Uses: G001, S009

S011 · calculation

Before

primitive coefficient at degree5 := div(0, 5)

After

primitive coefficient at degree5 = 0

Conditions: none

Since (x^(k+1))′=(k+1)x^k,divide the coefficient by the positive nonzero integer k+1.

Step identifiers and dependencies

Rule: exact_rational_arithmetic · Uses: G001, S010

S012 · calculation

Before

differentiate back coefficient4 := mul(0, 5)

After

differentiate back coefficient4 = 0

Conditions: none

Differentiation multiplies back by k+1 and must recover the original coefficient.

Step identifiers and dependencies

Rule: exact_rational_arithmetic · Uses: G001, S011

S013 · calculation

Before

primitive coefficient at degree6 := div(-18, 6)

After

primitive coefficient at degree6 = -3

Conditions: none

Since (x^(k+1))′=(k+1)x^k,divide the coefficient by the positive nonzero integer k+1.

Step identifiers and dependencies

Rule: exact_rational_arithmetic · Uses: G001, S012

S014 · calculation

Before

differentiate back coefficient5 := mul(-3, 6)

After

differentiate back coefficient5 = -18

Conditions: none

Differentiation multiplies back by k+1 and must recover the original coefficient.

Step identifiers and dependencies

Rule: exact_rational_arithmetic · Uses: G001, S013

S015 · implies

Before

[0, 0, 0, 0, 0, 0, -3]

After

all derivative coefficients agree with f

Conditions: none

Coefficient agreement proves the polynomial identity on all reals,not merely equality at a few points.

Step identifiers and dependencies

Rule: global_polynomial_identity · Uses: G001, S014

S016 · implies

Before

[0, 0, 0, 0, 0, 0, -3]

After

reduced coefficients;integer coefficients retain /1

Conditions: none

Return ascending coefficients,trim redundant leading-in-degree zeros,and retain[0/1] for the zero polynomial.

Step identifiers and dependencies

Rule: coefficient_output_protocol · Uses: G001, S015

Reverse checks

  • S016 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
  • S015 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
  • S014 · premise_and_result_audit — Recompute differentiate back coefficient5 with exact rational arithmetic, not rounded decimals.
  • S013 · premise_and_result_audit — Recompute primitive coefficient at degree6 with exact rational arithmetic, not rounded decimals.
  • S012 · premise_and_result_audit — Recompute differentiate back coefficient4 with exact rational arithmetic, not rounded decimals.
  • S011 · premise_and_result_audit — Recompute primitive coefficient at degree5 with exact rational arithmetic, not rounded decimals.
  • S010 · premise_and_result_audit — Recompute differentiate back coefficient3 with exact rational arithmetic, not rounded decimals.
  • S009 · premise_and_result_audit — Recompute primitive coefficient at degree4 with exact rational arithmetic, not rounded decimals.
  • S008 · premise_and_result_audit — Recompute differentiate back coefficient2 with exact rational arithmetic, not rounded decimals.
  • S007 · premise_and_result_audit — Recompute primitive coefficient at degree3 with exact rational arithmetic, not rounded decimals.
  • S006 · premise_and_result_audit — Recompute differentiate back coefficient1 with exact rational arithmetic, not rounded decimals.
  • S005 · premise_and_result_audit — Recompute primitive coefficient at degree2 with exact rational arithmetic, not rounded decimals.
  • S004 · premise_and_result_audit — Recompute differentiate back coefficient0 with exact rational arithmetic, not rounded decimals.
  • S003 · premise_and_result_audit — Recompute primitive coefficient at degree1 with exact rational arithmetic, not rounded decimals.
  • S002 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
  • S001 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.

Result

[0/1, 0/1, 0/1, 0/1, 0/1, 0/1, -3/1]

The full substitutions and premise checks give the original answer; source identity,format and status are unchanged.

This question review and source binding

Review date: 2026-10-26

Original questions, answers, step expressions, numeric ASTs and historical records are retained. Source candidate labels remain historical and are not overwritten as formal proof.

Source hashes and review receipts
  • input commit: 0f7be61e2c22317a4075561dc822c87d881a50db
  • original question SHA-256: c56215c556cc5794be163fa0bde3f54ac959f021d59cd93a92421f0299bf0656
  • candidate record SHA-256: 8e9d735a08ba7b36cba5412b3f09b946724a5e35243559c958d8c244f33eac6c
  • semantic review receipt SHA-256: c3b44ce8ed60ada6902fb206627537d5f90ba2a867eaba43a36f903280c496cc
  • question-first witness SHA-256: 3e91602adc7f81a2671a2a3ffaf6a447028b56abc5853e8271673d76f30cd7ad
  • Lean: NOT_RUN · human review: NOT_PERFORMED
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