{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-definite-integral-01",
  "mko_id": "mko-definite-integral",
  "title": {
    "zh-Hant": "定積分：以左端和的極限與矩公式計算 ∫₀³(x²−4)dx",
    "en": "Definite integral: computing ∫₀³(x²−4)dx as a limit of left sums and by the moment formula"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-definite-integral-01",
    "source": {
      "mko_id": "mko-definite-integral",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "f1fc56fc2ac48870dcaa53e6269bb2e365c1529c5540683ba6b6cd103ddcc5f7",
      "english_sha256": "cd470242748859a6d2bf765f69c47fa6a116eca1f2e64fb1a5edea6fe686d6a4"
    },
    "mode": "worked_example",
    "audience_prerequisites": [
      "finite_sum_rules",
      "seq_limit_def",
      "archimedean",
      "order_basics",
      "numeric_evaluation"
    ],
    "goal": {
      "zh-Hant": "設 f(x)=x²−4、a=0、b=3。(1) 只用 Riemann 定義（所有合法標記分割和在網格趨 0 時的共同極限）與 f 可積的前提，由 N 等分左端和 L_N 的極限求出 I(f,0,3)；(2) 用本 MKO 明列的矩公式獨立再算一次並核對；(3) 求反向端點的 I(f,3,0)；(4) 以 N=3、N=50 說明有限左端和不等於積分，並說明有號積分不是非負面積。",
      "en": "Let f(x)=x²−4, a=0, b=3. (1) Using only the Riemann definition (the common limit of all valid tagged-partition sums as the mesh tends to zero) and the premise that f is integrable, find I(f,0,3) as the limit of the N-piece left sums L_N; (2) recompute it independently with the moment formula listed in this MKO and compare; (3) find the reversed integral I(f,3,0); (4) use N=3 and N=50 to show that a finite left sum is not the integral, and explain why the signed integral is not a nonnegative area."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "f(x)=x²−4，端點 a=0≤b=3。f 是整數係數、次數 2 的多項式（升冪係數 −4、0、1），符合 MKO 假設 a1。",
          "en": "f(x)=x²−4 with endpoints a=0≤b=3. f is a polynomial of degree 2 with integer coefficients (ascending coefficients −4, 0, 1), as MKO assumption a1 requires."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "MKO 假設 a1：f 在全 ℝ 連續，並在有限區間上可積；所以 I(f,0,3) 存在，是一個實數。",
          "en": "MKO assumption a1: f is continuous on all of ℝ and integrable on finite intervals; so I(f,0,3) exists and is a real number."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "MKO 假設 a3 的左端和：對 a≤b 與正整數 N，Δx=(b−a)/N，x_i=a+iΔx，L_N=Σ_{i=0}^{N−1} f(x_i)Δx。本例 Δx=3/N、x_i=3i/N。",
          "en": "The left sum of MKO assumption a3: for a≤b and a positive integer N, Δx=(b−a)/N, x_i=a+iΔx and L_N=Σ_{i=0}^{N−1} f(x_i)Δx. Here Δx=3/N and x_i=3i/N."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1",
          "G2"
        ],
        "before_tex": "I(f,0,3)",
        "after_tex": "\\forall\\varepsilon>0\\ \\exists\\delta>0:\\ \\operatorname{mesh}(P)<\\delta\\Rightarrow|S(f,P,\\xi)-I(f,0,3)|<\\varepsilon",
        "relation": "definition",
        "rule": "mko_riemann_def",
        "conditions": [
          "0\\le 3"
        ],
        "explanation": {
          "zh-Hant": "依 MKO 的定義，I(f,0,3) 是 [0,3] 上所有合法標記分割和在網格趨 0 時的共同極限；依 riemann_limit_meaning，這句話的精確意思是：對每個 ε>0 都有 δ>0，使每個網格小於 δ 的標記分割 (P,ξ) 都有 |S(f,P,ξ)−I(f,0,3)|<ε。G2 保證這樣的數存在；本稿的工作是求出它的值。",
          "en": "By the MKO's definition, I(f,0,3) is the common limit of all valid tagged-partition sums on [0,3] as the mesh tends to zero; by riemann_limit_meaning this means precisely: for every ε>0 there is δ>0 such that every tagged partition (P,ξ) with mesh less than δ has |S(f,P,ξ)−I(f,0,3)|<ε. G2 guarantees that such a number exists; the task here is to find its value."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "G2",
            "mko_riemann_def",
            "riemann_limit_meaning"
          ],
          "explanation": {
            "zh-Hant": "此步只展開定義並記下存在性前提，不推出任何數值。存在性來自 G2（MKO 假設 a1），不是由後面某一族特殊的和收斂推出的。",
            "en": "This step only unfolds the definition and records the existence premise; it derives no value. Existence comes from G2 (MKO assumption a1), not from the convergence of one special family of sums later on."
          }
        },
        "leaf_basis": "mko_riemann_def + riemann_limit_meaning + def_unfold"
      },
      {
        "id": "S02",
        "uses": [
          "G3",
          "S01"
        ],
        "before_tex": "L_N=\\sum_{i=0}^{N-1} f(x_i)\\,\\Delta x",
        "after_tex": "L_N=S(f,P_N,\\xi_N),\\quad \\operatorname{mesh}(P_N)=\\tfrac{3}{N}",
        "relation": "implies",
        "rule": "riemann_limit_meaning",
        "conditions": [
          "N\\ge 1"
        ],
        "explanation": {
          "zh-Hant": "取分點 x_i=3i/N（i=0 到 N）。相鄰兩點相差 3/N>0，所以分點嚴格遞增，且 x_0=0、x_N=3，構成 [0,3] 的分割 P_N。第 i+1 段 [x_i,x_{i+1}] 的標記取左端點 x_i，它確實落在該段內。每段長度都是 3/N，所以網格是 3/N。於是 L_N 正是一個合法的標記分割和。",
          "en": "Take the division points x_i=3i/N for i=0 to N. Consecutive points differ by 3/N>0, so they increase strictly, with x_0=0 and x_N=3; they form a partition P_N of [0,3]. Piece i+1, [x_i,x_{i+1}], is tagged by its left endpoint x_i, which does lie in that piece. Every piece has length 3/N, so the mesh is 3/N. Thus L_N is exactly a valid tagged-partition sum."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G3",
            "riemann_limit_meaning",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "需要 N≥1，Δx=3/N 才有定義且為正。只宣稱每個左端和是標記分割和；反向（每個標記分割和都是左端和）並不成立，也不宣稱。",
            "en": "It needs N≥1, so that Δx=3/N is defined and positive. It claims only that each left sum is a tagged-partition sum; the converse (every tagged-partition sum is a left sum) is false and is not claimed."
          }
        },
        "leaf_basis": "riemann_limit_meaning + order_basics + numeric_evaluation"
      },
      {
        "id": "S03",
        "uses": [
          "S01",
          "S02",
          "archimedean"
        ],
        "before_tex": "\\operatorname{mesh}(P_N)=\\tfrac{3}{N}\\ \\text{and S01}",
        "after_tex": "L_N\\to I(f,0,3)\\quad(N\\to\\infty)",
        "relation": "implies",
        "rule": "seq_limit_def",
        "conditions": [
          "N\\ge 1"
        ],
        "explanation": {
          "zh-Hant": "給定 ε>0，取 S01 的 δ。依阿基米德性質取自然數 N_0>3/δ（且 N_0≥1）。對每個 N≥N_0，3/N≤3/N_0<δ，所以 P_N 的網格小於 δ，由 S01 得 |L_N−I(f,0,3)|<ε。這正是數列極限 L_N→I(f,0,3) 的定義。",
          "en": "Given ε>0, take the δ of S01. By the Archimedean property choose a natural number N_0>3/δ (with N_0≥1). For every N≥N_0, 3/N≤3/N_0<δ, so the mesh of P_N is below δ and S01 gives |L_N−I(f,0,3)|<ε. This is exactly the definition of the sequence limit L_N→I(f,0,3)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S01",
            "S02",
            "G2",
            "seq_limit_def",
            "archimedean"
          ],
          "explanation": {
            "zh-Hant": "這一步用到 I(f,0,3) 的存在（G2）。反向不宣稱：單看一族左端和收斂，不能推出函數可積；MKO 段落 0 也提醒，定義要求所有合法標記分割的共同極限。",
            "en": "This step uses the existence of I(f,0,3) (G2). The converse is not claimed: convergence of one family of left sums does not imply integrability; MKO paragraph 0 also stresses that the definition requires the common limit over all valid tagged partitions."
          }
        },
        "leaf_basis": "seq_limit_def + archimedean + order_basics"
      },
      {
        "id": "S04",
        "uses": [
          "G1",
          "G3"
        ],
        "before_tex": "f(x_i)\\,\\Delta x",
        "after_tex": "\\tfrac{27i^2}{N^3}-\\tfrac{12}{N}",
        "relation": "equivalent",
        "rule": "distributive",
        "conditions": [
          "N\\ge 1"
        ],
        "explanation": {
          "zh-Hant": "代入 x_i=3i/N：f(x_i)=(3i/N)²−4=9i²/N²−4；再乘 Δx=3/N，用分配律得 27i²/N³−12/N。",
          "en": "Substitute x_i=3i/N: f(x_i)=(3i/N)²−4=9i²/N²−4; multiplying by Δx=3/N and distributing gives 27i²/N³−12/N."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "distributive",
            "frac_rules",
            "power_laws"
          ],
          "explanation": {
            "zh-Hant": "反向：提出公因子 3/N，得 (3/N)(9i²/N²−4)=f(x_i)Δx。N≠0 時兩式是同一個數。",
            "en": "Backward: factor out 3/N to get (3/N)(9i²/N²−4)=f(x_i)Δx. For N≠0 both expressions are the same number."
          }
        },
        "leaf_basis": "distributive + frac_rules + power_laws"
      },
      {
        "id": "S05",
        "uses": [
          "S04",
          "finite_sum_rules"
        ],
        "before_tex": "L_N=\\sum_{i=0}^{N-1}\\left(\\tfrac{27i^2}{N^3}-\\tfrac{12}{N}\\right)",
        "after_tex": "L_N=\\tfrac{27}{N^3}\\sum_{i=0}^{N-1} i^2-12",
        "relation": "equivalent",
        "rule": "finite_sum_rules",
        "conditions": [
          "N\\ge 1"
        ],
        "explanation": {
          "zh-Hant": "有限和可逐項拆開，常數因子 27/N³ 可提出和號外；常數項 12/N 在 i=0 到 N−1 共出現 N 次，和為 N·(12/N)=12。",
          "en": "A finite sum splits term by term and the constant factor 27/N³ comes out of the sum; the constant term 12/N appears N times (i=0 to N−1), giving N·(12/N)=12."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "finite_sum_rules",
            "frac_rules"
          ],
          "explanation": {
            "zh-Hant": "反向：把 27/N³ 乘回每一項、把 12 寫成 N 個 12/N 之和，就回到原式；所用的有限和規則都是恆等式。",
            "en": "Backward: multiply 27/N³ back into each term and write 12 as the sum of N copies of 12/N to return to the original; the finite-sum rules used are identities."
          }
        },
        "leaf_basis": "finite_sum_rules + frac_rules"
      },
      {
        "id": "S06",
        "uses": [
          "S05",
          "L_sum_sq"
        ],
        "before_tex": "\\sum_{i=0}^{N-1} i^2",
        "after_tex": "\\tfrac{(N-1)N(2N-1)}{6}",
        "relation": "equivalent",
        "rule": "L_sum_sq",
        "conditions": [
          "N\\ge 1"
        ],
        "explanation": {
          "zh-Hant": "依引理 L_sum_sq（在 S06.1 到 S06.3 證明）把平方和換成封閉式。",
          "en": "By Lemma L_sum_sq (proved in S06.1 to S06.3), replace the sum of squares by its closed form."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_sum_sq"
          ],
          "explanation": {
            "zh-Hant": "兩式對每個 N≥1 相等，所以替換雙向成立。",
            "en": "The two expressions are equal for every N≥1, so the replacement works in both directions."
          }
        },
        "leaf_basis": "nat_induction + finite_sum_rules + distributive"
      },
      {
        "id": "S06.1",
        "uses": [
          "finite_sum_rules"
        ],
        "before_tex": "\\sum_{i=0}^{0} i^2",
        "after_tex": "0=\\tfrac{(1-1)\\cdot 1\\cdot(2\\cdot 1-1)}{6}",
        "relation": "induction_base",
        "rule": "numeric_evaluation",
        "conditions": [
          "N=1"
        ],
        "explanation": {
          "zh-Hant": "N=1：和只有 i=0 一項，其值 0²=0；右邊 (1−1)·1·(2−1)/6=0。基礎情形成立。",
          "en": "N=1: the sum has the single term i=0, equal to 0²=0; the right side is (1−1)·1·(2−1)/6=0. The base case holds."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "finite_sum_rules",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "只需直接計算兩邊（N2）。",
            "en": "Only direct evaluation of both sides is needed (N2)."
          }
        },
        "leaf_basis": "finite_sum_rules + numeric_evaluation"
      },
      {
        "id": "S06.2",
        "uses": [
          "finite_sum_rules",
          "distributive"
        ],
        "before_tex": "\\sum_{i=0}^{N-1} i^2=\\tfrac{(N-1)N(2N-1)}{6}",
        "after_tex": "\\sum_{i=0}^{N} i^2=\\tfrac{N(N+1)(2N+1)}{6}",
        "relation": "induction_step",
        "rule": "finite_sum_rules",
        "conditions": [
          "N\\ge 1"
        ],
        "explanation": {
          "zh-Hant": "假設 N 的公式成立。新的和多了一項 i=N：Σ_{i=0}^{N} i²=(N−1)N(2N−1)/6+N²=N[(N−1)(2N−1)+6N]/6=N(2N²+3N+1)/6=N(N+1)(2N+1)/6。這正是把 N 換成 N+1 的公式，因為 (N+1)−1=N、2(N+1)−1=2N+1。",
          "en": "Assume the formula for N. The new sum has one more term, i=N: Σ_{i=0}^{N} i²=(N−1)N(2N−1)/6+N²=N[(N−1)(2N−1)+6N]/6=N(2N²+3N+1)/6=N(N+1)(2N+1)/6. This is the formula with N replaced by N+1, since (N+1)−1=N and 2(N+1)−1=2N+1."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "finite_sum_rules",
            "distributive"
          ],
          "explanation": {
            "zh-Hant": "只用到 N 的歸納假設與多項式恆等式 (N−1)N(2N−1)/6+N²=N(N+1)(2N+1)/6（N1，次數 3 的格點恆等式核對）。",
            "en": "It uses only the induction hypothesis for N and the polynomial identity (N−1)N(2N−1)/6+N²=N(N+1)(2N+1)/6 (N1, a grid check of a degree-3 identity)."
          }
        },
        "leaf_basis": "finite_sum_rules + distributive + add_comm"
      },
      {
        "id": "S06.3",
        "uses": [
          "S06.1",
          "S06.2",
          "nat_induction"
        ],
        "before_tex": "P(1)\\ \\land\\ \\forall N\\ge 1\\,(P(N)\\Rightarrow P(N+1))",
        "after_tex": "\\forall N\\ge 1:\\ P(N)",
        "relation": "implies",
        "rule": "nat_induction",
        "conditions": [],
        "explanation": {
          "zh-Hant": "令 P(N) 為「Σ_{i=0}^{N−1} i²=(N−1)N(2N−1)/6」。S06.1 給 P(1)，S06.2 給 P(N)⇒P(N+1)；依數學歸納原理（n₀=1），P(N) 對每個 N≥1 成立。S06.1 到 S06.3 合起來證得引理 L_sum_sq。",
          "en": "Let P(N) be 'Σ_{i=0}^{N−1} i²=(N−1)N(2N−1)/6'. S06.1 gives P(1) and S06.2 gives P(N)⇒P(N+1); by the induction principle (n₀=1), P(N) holds for every N≥1. S06.1 to S06.3 together prove Lemma L_sum_sq."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S06.1",
            "S06.2",
            "nat_induction"
          ],
          "explanation": {
            "zh-Hant": "歸納原理需要基礎與歸納步兩者；兩者都已證，且證明中沒有使用引理本身。",
            "en": "Induction needs both the base and the step; both are proved, and neither uses the lemma itself."
          }
        },
        "leaf_basis": "nat_induction"
      },
      {
        "id": "S07",
        "uses": [
          "S05",
          "S06"
        ],
        "before_tex": "L_N=\\tfrac{27}{N^3}\\cdot\\tfrac{(N-1)N(2N-1)}{6}-12",
        "after_tex": "L_N=-3-\\tfrac{27}{2N}+\\tfrac{9}{2N^2}",
        "relation": "equivalent",
        "rule": "frac_rules",
        "conditions": [
          "N\\ge 1"
        ],
        "explanation": {
          "zh-Hant": "約分：27/N³·(N−1)N(2N−1)/6=9(N−1)(2N−1)/(2N²)=9(2N²−3N+1)/(2N²)=9−27/(2N)+9/(2N²)。再減 12 得 −3−27/(2N)+9/(2N²)。約去 N 需要 N≠0，由 N≥1 保證。",
          "en": "Simplify: 27/N³·(N−1)N(2N−1)/6=9(N−1)(2N−1)/(2N²)=9(2N²−3N+1)/(2N²)=9−27/(2N)+9/(2N²). Subtracting 12 gives −3−27/(2N)+9/(2N²). Cancelling N needs N≠0, which N≥1 guarantees."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "frac_rules",
            "distributive",
            "mul_inverse_nonzero"
          ],
          "explanation": {
            "zh-Hant": "兩邊同乘非零的 2N² 都化成同一個多項式 −6N²−27N+9（N3 為次數 2 的恆等式核對）；再除回 2N² 是允許的，所以雙向成立。N4 另在 N=1 到 50 逐點核對，那只是有限核對。",
            "en": "Multiplying both sides by the nonzero 2N² turns each into the same polynomial −6N²−27N+9 (N3 checks this degree-2 identity); dividing back by 2N² is allowed, so the step is two-way. N4 also checks N=1 to 50 point by point, which is only a finite check."
          }
        },
        "leaf_basis": "frac_rules + distributive + mul_inverse_nonzero"
      },
      {
        "id": "S08",
        "uses": [
          "S07",
          "G3"
        ],
        "before_tex": "L_3,\\ L_{50}",
        "after_tex": "L_3=-7,\\quad L_{50}=-\\tfrac{16341}{5000}",
        "relation": "implies",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "N=3：Δx=1，左端點 0、1、2，f 值 −4、−3、0，直接相加得 L_3=−7；公式給 −3−27/6+9/18=−3−9/2+1/2=−7，一致。N=50（MKO 假設 a3 允許的最大 N）：L_50=−3−27/100+9/5000=−16341/5000=−3.2682。兩者都不等於 −3。",
          "en": "N=3: Δx=1, left points 0, 1, 2, values of f −4, −3, 0, so directly L_3=−7; the formula gives −3−27/6+9/18=−3−9/2+1/2=−7, which agrees. N=50 (the largest N allowed by MKO assumption a3): L_50=−3−27/100+9/5000=−16341/5000=−3.2682. Neither equals −3."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S07",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "這是兩個 N 的數值（N5 到 N8），用來對照 S07 的公式並顯示有限和不是積分；它們不證明極限。",
            "en": "These are values at two N (N5 to N8), used to cross-check the formula of S07 and to show that a finite sum is not the integral; they do not prove the limit."
          }
        },
        "leaf_basis": "numeric_evaluation"
      },
      {
        "id": "S09",
        "uses": [
          "S07"
        ],
        "before_tex": "|L_N-(-3)|",
        "after_tex": "\\tfrac{27N-9}{2N^2}\\le\\tfrac{27}{2N}",
        "relation": "implies",
        "rule": "abs_rules",
        "conditions": [
          "N\\ge 1"
        ],
        "explanation": {
          "zh-Hant": "由 S07，L_N+3=−27/(2N)+9/(2N²)=−(27N−9)/(2N²)。N≥1 時 27N−9≥18>0，所以 |L_N+3|=(27N−9)/(2N²)。又 −9<0，故 (27N−9)/(2N²)≤27N/(2N²)=27/(2N)。",
          "en": "By S07, L_N+3=−27/(2N)+9/(2N²)=−(27N−9)/(2N²). For N≥1, 27N−9≥18>0, so |L_N+3|=(27N−9)/(2N²). Since −9<0, (27N−9)/(2N²)≤27N/(2N²)=27/(2N)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S07",
            "abs_rules",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "需要 N≥1 才能確定 27N−9>0 並去掉絕對值；不等式只是單向上界，不宣稱相等。N10 在 N=1 到 50 做有限核對。",
            "en": "N≥1 is needed to know 27N−9>0 and remove the absolute value; the inequality is a one-way bound, not an equality. N10 is a finite check for N=1 to 50."
          }
        },
        "leaf_basis": "abs_rules + order_basics + frac_rules"
      },
      {
        "id": "S10",
        "uses": [
          "S09",
          "archimedean"
        ],
        "before_tex": "|L_N+3|\\le\\tfrac{27}{2N}\\ \\ (N\\ge 1)",
        "after_tex": "L_N\\to -3",
        "relation": "implies",
        "rule": "seq_limit_def",
        "conditions": [
          "\\varepsilon>0"
        ],
        "explanation": {
          "zh-Hant": "給定 ε>0，依阿基米德性質取自然數 N_1>27/(2ε)。對每個 N≥N_1，|L_N−(−3)|≤27/(2N)≤27/(2N_1)<ε。依數列極限的定義，L_N→−3。",
          "en": "Given ε>0, the Archimedean property gives a natural number N_1>27/(2ε). For every N≥N_1, |L_N−(−3)|≤27/(2N)≤27/(2N_1)<ε. By the definition of a sequence limit, L_N→−3."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S09",
            "seq_limit_def",
            "archimedean"
          ],
          "explanation": {
            "zh-Hant": "定義要求對所有 N≥N_1 成立，不只 N≤50；因此 S09 的上界必須對一般 N 證明，不能用有限表格代替。",
            "en": "The definition requires the bound for all N≥N_1, not only N≤50; this is why the bound of S09 had to be proved for general N rather than read off a finite table."
          }
        },
        "leaf_basis": "seq_limit_def + archimedean + order_basics"
      },
      {
        "id": "S11",
        "uses": [
          "S03",
          "S10",
          "limit_unique"
        ],
        "before_tex": "L_N\\to I(f,0,3),\\quad L_N\\to -3",
        "after_tex": "I(f,0,3)=-3",
        "relation": "implies",
        "rule": "limit_unique",
        "conditions": [],
        "explanation": {
          "zh-Hant": "同一個數列 L_N 既收斂到 I(f,0,3)（S03），又收斂到 −3（S10）；數列極限唯一，所以 I(f,0,3)=−3。",
          "en": "The same sequence L_N converges to I(f,0,3) (S03) and to −3 (S10); a sequence has at most one limit, so I(f,0,3)=−3."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S03",
            "S10",
            "G2",
            "limit_unique"
          ],
          "explanation": {
            "zh-Hant": "需要兩個極限屬於同一數列；S03 又依賴 G2 的存在性。若不知 f 可積，只能得到「左端和的極限是 −3」，不能稱為積分。",
            "en": "Both limits must concern the same sequence, and S03 relies on existence from G2. Without integrability one would only know that the left sums tend to −3, which could not be called the integral."
          }
        },
        "leaf_basis": "limit_unique + modus_ponens"
      },
      {
        "id": "S12",
        "uses": [
          "G1",
          "mko_moment"
        ],
        "before_tex": "I(f,0,3)=I(x^2,0,3)-4\\,I(x^0,0,3)",
        "after_tex": "\\tfrac{3^3-0^3}{3}-4\\cdot\\tfrac{3^1-0^1}{1}=9-12=-3",
        "relation": "theorem_application",
        "rule": "mko_moment",
        "conditions": [
          "0\\le 3"
        ],
        "explanation": {
          "zh-Hant": "第二條路：MKO 明列的矩公式逐係數相加。x² 的矩是 (3³−0³)/3=9，常數 1=x⁰ 的矩是 (3−0)/1=3，所以 I(f,0,3)=9−4·3=−3，與 S11 相同。",
          "en": "A second route: the MKO's listed moment formula, summed coefficientwise. The moment of x² is (3³−0³)/3=9 and that of the constant 1=x⁰ is (3−0)/1=3, so I(f,0,3)=9−4·3=−3, the same as S11."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "mko_moment",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "矩公式是本切片明列的規則，此處只是套用；兩條路一致是互相核對。這不是一般微積分基本定理的證明，也不宣稱是。",
            "en": "The moment formula is a rule listed for this slice and is only applied here; agreement of the two routes is a cross-check. It is not, and is not claimed to be, a proof of the general fundamental theorem of calculus."
          }
        },
        "leaf_basis": "mko_moment + numeric_evaluation"
      },
      {
        "id": "S13",
        "uses": [
          "S11",
          "mko_reverse"
        ],
        "before_tex": "I(f,3,0)",
        "after_tex": "-I(f,0,3)=3",
        "relation": "equivalent",
        "rule": "mko_reverse",
        "conditions": [],
        "explanation": {
          "zh-Hant": "反向端點要換號：I(f,3,0)=−I(f,0,3)=−(−3)=3。",
          "en": "Reversing the endpoints changes the sign: I(f,3,0)=−I(f,0,3)=−(−3)=3."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "S11",
            "mko_reverse"
          ],
          "explanation": {
            "zh-Hant": "再反向一次：I(f,0,3)=−I(f,3,0)=−3，回到 S11；換號規則兩向都成立。",
            "en": "Reversing once more gives I(f,0,3)=−I(f,3,0)=−3, back to S11; the sign rule works in both directions."
          }
        },
        "leaf_basis": "mko_reverse + neg_mul"
      },
      {
        "id": "S14",
        "uses": [
          "G1",
          "mko_moment"
        ],
        "before_tex": "I(f,0,2),\\ I(f,2,3)",
        "after_tex": "-\\tfrac{16}{3},\\ \\tfrac{7}{3}",
        "relation": "theorem_application",
        "rule": "mko_moment",
        "conditions": [
          "0\\le 2",
          "2\\le 3"
        ],
        "explanation": {
          "zh-Hant": "用矩公式分段：I(f,0,2)=(2³−0)/3−4·2=8/3−8=−16/3；I(f,2,3)=(3³−2³)/3−4·(3−2)=19/3−4=7/3。兩段相加 −16/3+7/3=−3=I(f,0,3)。符號的來由：f(x)=(x−2)(x+2)，在 0≤x<2 時為負（例如 f(1)=−3），在 2<x≤3 時為正。",
          "en": "Apply the moment formula piecewise: I(f,0,2)=(2³−0)/3−4·2=8/3−8=−16/3; I(f,2,3)=(3³−2³)/3−4·(3−2)=19/3−4=7/3. The two pieces add to −16/3+7/3=−3=I(f,0,3). The signs come from f(x)=(x−2)(x+2), which is negative for 0≤x<2 (for example f(1)=−3) and positive for 2<x≤3."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "mko_moment",
            "numeric_evaluation",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "只是矩公式在兩段上的套用（N12 到 N14、N16）；不宣稱一般函數的區間可加性。",
            "en": "This only applies the moment formula on two pieces (N12 to N14, N16); additivity over intervals for general functions is not claimed."
          }
        },
        "leaf_basis": "mko_moment + numeric_evaluation + order_basics"
      },
      {
        "id": "S15",
        "uses": [
          "S11",
          "S14"
        ],
        "before_tex": "|I(f,0,3)|",
        "after_tex": "3\\neq\\tfrac{23}{3}=|I(f,0,2)|+|I(f,2,3)|",
        "relation": "implies",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "有號積分讓負的一段與正的一段互相抵消：|I(f,0,3)|=3，而兩段絕對值之和是 16/3+7/3=23/3。所以「積分就是非負面積」或「取絕對值就得到積分」在這裡都不成立；MKO 假設 a2 明說積分是有向、有號的。",
          "en": "The signed integral lets the negative piece and the positive piece cancel: |I(f,0,3)|=3, while the absolute values of the two pieces add to 16/3+7/3=23/3. So 'the integral is a nonnegative area' and 'taking an absolute value gives the integral' both fail here; MKO assumption a2 states that the integral is oriented and signed."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S11",
            "S14",
            "abs_rules"
          ],
          "explanation": {
            "zh-Hant": "這是具體數字的比較（N15）。本稿不發展面積理論，23/3 只是兩段絕對值之和。",
            "en": "This compares explicit numbers (N15). Area theory is not developed here; 23/3 is only the sum of the two absolute values."
          }
        },
        "leaf_basis": "abs_rules + numeric_evaluation"
      },
      {
        "id": "S16",
        "uses": [
          "S08",
          "S11",
          "S12",
          "S13",
          "S15"
        ],
        "before_tex": "\\text{goals (1) to (4)}",
        "after_tex": "I(f,0,3)=-3,\\ I(f,3,0)=3,\\ L_3=-7,\\ L_{50}=-\\tfrac{16341}{5000}",
        "relation": "implies",
        "rule": "and_intro_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "合併：(1) S11 由定義得 I(f,0,3)=−3；(2) S12 以矩公式得相同值；(3) S13 得 I(f,3,0)=3；(4) S08 顯示有限左端和不等於積分，S15 顯示有號積分不是非負面積。",
          "en": "Combine: (1) S11 gives I(f,0,3)=−3 from the definition; (2) S12 gives the same value by the moment formula; (3) S13 gives I(f,3,0)=3; (4) S08 shows finite left sums differ from the integral and S15 shows the signed integral is not a nonnegative area."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S08",
            "S11",
            "S12",
            "S13",
            "S15",
            "and_intro_elim"
          ],
          "explanation": {
            "zh-Hant": "目標 G：四個小題。足夠條件 H：S08、S11、S12、S13、S15 的結論同時成立；H⇒G 只是合取。H 由 G1 到 G3 正向證成，沒有把 G 當作前提。",
            "en": "Goal G: the four parts. Sufficient condition H: the conclusions of S08, S11, S12, S13 and S15 together; H⇒G is just conjunction. H is proved forwards from G1 to G3; G is never used as a premise."
          }
        },
        "leaf_basis": "and_intro_elim"
      }
    ],
    "result": {
      "zh-Hant": "在 f 可積（MKO 假設 a1）的前提下，由 Riemann 定義與左端和 L_N=−3−27/(2N)+9/(2N²) 的極限得 ∫₀³(x²−4)dx=−3；MKO 的矩公式獨立給出同一值。反向端點 ∫₃⁰(x²−4)dx=3。有限左端和 L_3=−7、L_50=−3.2682 都不等於積分；兩段 −16/3 與 7/3 互相抵消，所以積分不是非負面積。",
      "en": "Given integrability of f (MKO assumption a1), the Riemann definition and the limit of the left sums L_N=−3−27/(2N)+9/(2N²) give ∫₀³(x²−4)dx=−3; the MKO's moment formula independently gives the same value. With reversed endpoints ∫₃⁰(x²−4)dx=3. The finite left sums L_3=−7 and L_50=−3.2682 differ from the integral; the pieces −16/3 and 7/3 cancel, so the integral is not a nonnegative area."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "L_50=−3.2682 就是 ∫₀³(x²−4)dx",
          "en": "L_50=−3.2682 is ∫₀³(x²−4)dx"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "L_50 與 −3 相差 (27·50−9)/(2·50²)=1341/5000≠0；積分是 N→∞ 的極限。",
          "en": "L_50 differs from −3 by (27·50−9)/(2·50²)=1341/5000≠0; the integral is the limit as N→∞."
        }
      },
      {
        "claim": {
          "zh-Hant": "∫₃⁰(x²−4)dx=∫₀³(x²−4)dx=−3，因為是同一段區間",
          "en": "∫₃⁰(x²−4)dx=∫₀³(x²−4)dx=−3 because it is the same interval"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "反向規則給 I(f,3,0)=−I(f,0,3)=3。",
          "en": "The reversal rule gives I(f,3,0)=−I(f,0,3)=3."
        }
      },
      {
        "claim": {
          "zh-Hant": "定積分恆非負，所以 ∫₀³(x²−4)dx 應等於 23/3",
          "en": "A definite integral is never negative, so ∫₀³(x²−4)dx should be 23/3"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "積分有號，值為 −3；23/3 是兩段絕對值之和。",
          "en": "The integral is signed and equals −3; 23/3 is the sum of the absolute values of the two pieces."
        }
      },
      {
        "claim": {
          "zh-Hant": "只要某一族等分左端和收斂，函數就可積，積分就是那個極限",
          "en": "If one family of equal-piece left sums converges, the function is integrable and the integral is that limit"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "Dirichlet 函數 D（有理數處為 1、其餘為 0）在 [0,1] 上：左端點 i/N 都是有理數，所以每個左端和都是 1；但改取無理數標記，網格再小的和也都是 0，沒有共同極限，D 不是 Riemann 可積（MKO 段落 0）。",
          "en": "The Dirichlet function D (1 at rationals, 0 elsewhere) on [0,1]: the left points i/N are rational, so every left sum is 1; but with irrational tags the sums are 0 for arbitrarily small mesh, so there is no common limit and D is not Riemann integrable (MKO paragraph 0)."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "id": "finite_sum_rules",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "有限和：Σ_{i=1}^{0} a_i=0、Σ_{i=1}^{n+1} a_i=Σ_{i=1}^{n} a_i + a_{n+1}；Σ(a_i+b_i)=Σa_i+Σb_i；Σ c·a_i=c·Σa_i。",
          "en": "Finite sums: Σ_{i=1}^{0} a_i=0 and Σ_{i=1}^{n+1} a_i=Σ_{i=1}^{n} a_i + a_{n+1}; Σ(a_i+b_i)=Σa_i+Σb_i; Σ c·a_i=c·Σa_i."
        }
      },
      {
        "id": "seq_limit_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "數列極限：a_n→L 表示對每個 ε>0，存在 N 使所有 n≥N 都有 |a_n−L|<ε。",
          "en": "Limit of a sequence: a_n→L means that for every ε>0 there is N such that |a_n−L|<ε for all n≥N."
        }
      },
      {
        "id": "archimedean",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "阿基米德性質：對任意實數 x，存在自然數 n>x；因此對任意 ε>0，存在 n 使 1/n<ε。",
          "en": "Archimedean property: for every real x there is a natural number n>x; hence for every ε>0 there is n with 1/n<ε."
        }
      },
      {
        "id": "order_basics",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "實數的序：任兩數恰有 u<v、u=v、u>v 之一；u<v 且 v<w 推出 u<w；u<v 推出 u+w<v+w；u<v 且 w>0 推出 uw<vw；u<v 且 w<0 推出 uw>vw。",
          "en": "Order of the reals: exactly one of u<v, u=v, u>v holds; u<v and v<w give u<w; u<v gives u+w<v+w; u<v and w>0 give uw<vw; u<v and w<0 give uw>vw."
        }
      },
      {
        "id": "numeric_evaluation",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "明確整數或有理數算式的精確求值；可以獨立重算。",
          "en": "Exact evaluation of an explicit integer or rational expression; independently recomputable."
        }
      },
      {
        "id": "def_unfold",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "展開定義：把一個已定義的詞換成它的定義內容（或反向收回），命題的真假不變。",
          "en": "Unfolding a definition: replacing a defined term by its definition (or folding it back) does not change whether a statement is true."
        }
      },
      {
        "id": "distributive",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "分配律：u(v+w)=uv+uw；配合乘法交換律也有 (v+w)u=vu+wu。",
          "en": "Distributivity: u(v+w)=uv+uw; with commutativity also (v+w)u=vu+wu."
        }
      },
      {
        "id": "frac_rules",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "分數運算（分母皆非零）：a/c+b/c=(a+b)/c；(a/b)(c/d)=(ac)/(bd)；(a/b)/(c/d)=(ad)/(bc)（另需 c≠0）；a/b=(ak)/(bk)（k≠0）。",
          "en": "Fraction rules (all denominators nonzero): a/c+b/c=(a+b)/c; (a/b)(c/d)=(ac)/(bd); (a/b)/(c/d)=(ad)/(bc) (also needing c≠0); a/b=(ak)/(bk) for k≠0."
        }
      },
      {
        "id": "power_laws",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "整數指數律（底數非零或指數為正整數）：a^m·a^n=a^{m+n}，(a^m)^n=a^{mn}，(ab)^n=a^n b^n，a^0=1。",
          "en": "Integer exponent laws (base nonzero, or exponents positive integers): a^m·a^n=a^{m+n}, (a^m)^n=a^{mn}, (ab)^n=a^n b^n, a^0=1."
        }
      },
      {
        "id": "nat_induction",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "數學歸納原理：對從 n₀ 起的命題族 P(n)，若 P(n₀) 成立，且對每個 n≥n₀ 都有 P(n)⇒P(n+1)，則所有 n≥n₀ 都有 P(n)。",
          "en": "Principle of induction: for a family P(n) starting at n₀, if P(n₀) holds and P(n)⇒P(n+1) for every n≥n₀, then P(n) holds for every n≥n₀."
        }
      },
      {
        "id": "add_comm",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "加法交換律：u+v=v+u。",
          "en": "Commutativity of addition: u+v=v+u."
        }
      },
      {
        "id": "mul_inverse_nonzero",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "非零數的倒數：在 ℚ、ℝ、ℂ 中，若 u≠0，存在 1/u 使 u·(1/u)=1；對 u≠0 定義 v/u 為 v·(1/u)。u=0 時沒有倒數，v/0 不定義。",
          "en": "Inverse of a nonzero number: in ℚ, ℝ or ℂ, if u≠0 there is 1/u with u·(1/u)=1, and v/u means v·(1/u). 0 has no inverse, so v/0 is undefined."
        }
      },
      {
        "id": "abs_rules",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "絕對值（實數）：|u|=u（u≥0）、|u|=−u（u<0）；|u|≥0；|uv|=|u||v|；三角不等式 |u+v|≤|u|+|v|；|u|<r ⇔ −r<u<r。",
          "en": "Absolute value (reals): |u|=u if u≥0 and |u|=−u if u<0; |u|≥0; |uv|=|u||v|; triangle inequality |u+v|≤|u|+|v|; |u|<r ⇔ −r<u<r."
        }
      },
      {
        "id": "modus_ponens",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "肯定前件：由 P 與 P⇒Q 得 Q。",
          "en": "Modus ponens: from P and P⇒Q infer Q."
        }
      },
      {
        "id": "neg_mul",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "負號與乘法：(−u)v=−(uv)，(−u)(−v)=uv。",
          "en": "Signs and multiplication: (−u)v=−(uv) and (−u)(−v)=uv."
        }
      },
      {
        "id": "and_intro_elim",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "合取：由 P 與 Q 得 P∧Q；由 P∧Q 可分別得 P 與 Q。",
          "en": "Conjunction: from P and Q infer P∧Q; from P∧Q infer P and infer Q."
        }
      },
      {
        "id": "mko_riemann_def",
        "kind": "cited",
        "source": {
          "mko_id": "mko-definite-integral",
          "sha256": "f1fc56fc2ac48870dcaa53e6269bb2e365c1529c5540683ba6b6cd103ddcc5f7",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "對有限實區間 a≤b 上的連續多項式 f，Riemann 定積分是所有合法標記分割和在網格趨 0 時的共同極限。",
          "en": "For a continuous polynomial f on a finite real interval a≤b, its Riemann integral is the common limit of all valid tagged-partition sums as the mesh tends to zero."
        }
      },
      {
        "id": "mko_reverse",
        "kind": "cited",
        "source": {
          "mko_id": "mko-definite-integral",
          "sha256": "f1fc56fc2ac48870dcaa53e6269bb2e365c1529c5540683ba6b6cd103ddcc5f7",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "反向端點換號：I(f,a,b)=−I(f,b,a)；相同端點 I(f,a,a)=0。",
          "en": "Reversing endpoints changes the sign: I(f,a,b)=−I(f,b,a); equal endpoints give I(f,a,a)=0."
        }
      },
      {
        "id": "mko_moment",
        "kind": "cited",
        "source": {
          "mko_id": "mko-definite-integral",
          "sha256": "f1fc56fc2ac48870dcaa53e6269bb2e365c1529c5540683ba6b6cd103ddcc5f7",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "本切片明列矩公式 I(x^n,a,b)=(b^{n+1}−a^{n+1})/(n+1)；對 f=Σc_k x^k 逐係數相加：I(f,a,b)=Σc_k(b^{k+1}−a^{k+1})/(k+1)。",
          "en": "This slice explicitly uses the moment formula I(x^n,a,b)=(b^{n+1}−a^{n+1})/(n+1); for f=Σc_k x^k the moments are summed coefficientwise: I(f,a,b)=Σc_k(b^{k+1}−a^{k+1})/(k+1)."
        }
      },
      {
        "id": "riemann_limit_meaning",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "「共同極限為 I」的精確意思：對每個 ε>0，存在 δ>0，使 [a,b] 的每個網格小於 δ 的標記分割 (P,ξ) 都有 |S(f,P,ξ)−I|<ε。分割 P 是有限嚴格遞增的分點 t_0=a<t_1<t_2 直到 t_m=b；標記 ξ_j 落在 [t_{j−1},t_j]；網格是最大的 t_j−t_{j−1}；S(f,P,ξ)=Σ_{j=1}^{m} f(ξ_j)(t_j−t_{j−1})。",
          "en": "The precise meaning of 'the common limit is I': for every ε>0 there is δ>0 such that every tagged partition (P,ξ) of [a,b] with mesh less than δ has |S(f,P,ξ)−I|<ε. A partition P is a finite strictly increasing list of points t_0=a<t_1<t_2 up to t_m=b; a tag ξ_j lies in [t_{j−1},t_j]; the mesh is the largest t_j−t_{j−1}; S(f,P,ξ)=Σ_{j=1}^{m} f(ξ_j)(t_j−t_{j−1})."
        }
      },
      {
        "id": "limit_unique",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "數列極限唯一：若 s_N→L 且 s_N→L′，則 L=L′。",
          "en": "Limits of sequences are unique: if s_N→L and s_N→L′, then L=L′."
        }
      },
      {
        "id": "L_sum_sq",
        "kind": "lemma",
        "proved_in": [
          "S06.1",
          "S06.2",
          "S06.3"
        ],
        "statement": {
          "zh-Hant": "對每個正整數 N，Σ_{i=0}^{N−1} i²=(N−1)N(2N−1)/6。",
          "en": "For every positive integer N, Σ_{i=0}^{N−1} i²=(N−1)N(2N−1)/6."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S06.2",
        "vars": [
          "N"
        ],
        "degree": 3,
        "lhs": "(N-1)*N*(2*N-1)/6 + N**2",
        "rhs": "N*(N+1)*(2*N+1)/6"
      },
      {
        "id": "N2",
        "step": "S06.1",
        "lhs": "0**2",
        "rhs": "(1-1)*1*(2*1-1)/6"
      },
      {
        "id": "N3",
        "step": "S07",
        "vars": [
          "N"
        ],
        "degree": 2,
        "lhs": "9*(N-1)*(2*N-1) - 24*N**2",
        "rhs": "-6*N**2 - 27*N + 9"
      },
      {
        "id": "N4",
        "step": "S07",
        "vars": [
          "N"
        ],
        "range": [
          1,
          50
        ],
        "lhs": "(27/N**3)*((N-1)*N*(2*N-1)/6) - 12",
        "rhs": "-3 - 27/(2*N) + 9/(2*N**2)"
      },
      {
        "id": "N5",
        "step": "S08",
        "lhs": "(0**2-4)*1 + (1**2-4)*1 + (2**2-4)*1",
        "rhs": "-7"
      },
      {
        "id": "N6",
        "step": "S08",
        "lhs": "-3 - 27/(2*3) + 9/(2*3**2)",
        "rhs": "-7"
      },
      {
        "id": "N7",
        "step": "S08",
        "lhs": "-3 - 27/(2*50) + 9/(2*50**2)",
        "rhs": "-16341/5000"
      },
      {
        "id": "N8",
        "step": "S08",
        "lhs": "-16341/5000",
        "relation": "!=",
        "rhs": "-3"
      },
      {
        "id": "N9",
        "step": "S07",
        "lhs": "(0**2-4)*3",
        "rhs": "-3 - 27/2 + 9/2"
      },
      {
        "id": "N10",
        "step": "S09",
        "vars": [
          "N"
        ],
        "range": [
          1,
          50
        ],
        "lhs": "abs(-27/(2*N) + 9/(2*N**2))",
        "relation": "<=",
        "rhs": "27/(2*N)"
      },
      {
        "id": "N11",
        "step": "S12",
        "lhs": "(3**3-0**3)/3 - 4*(3-0)",
        "rhs": "-3"
      },
      {
        "id": "N12",
        "step": "S14",
        "lhs": "(2**3-0**3)/3 - 4*(2-0)",
        "rhs": "-16/3"
      },
      {
        "id": "N13",
        "step": "S14",
        "lhs": "(3**3-2**3)/3 - 4*(3-2)",
        "rhs": "7/3"
      },
      {
        "id": "N14",
        "step": "S14",
        "lhs": "-16/3 + 7/3",
        "rhs": "-3"
      },
      {
        "id": "N15",
        "step": "S15",
        "lhs": "abs(-16/3) + abs(7/3)",
        "rhs": "23/3"
      },
      {
        "id": "N16",
        "step": "S14",
        "lhs": "1**2-4",
        "rhs": "-3"
      },
      {
        "id": "N17",
        "lhs": "(27*50-9)/(2*50**2)",
        "rhs": "1341/5000"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "只處理 MKO 範圍內的一個連續多項式 f(x)=x²−4（整數係數、次數 2）在有限區間 [0,3] 上的有號 Riemann 積分。f 的可積性取自 MKO 假設 a1，本稿不證明；本稿證明的是：在可積前提下，積分值必為 −3。不使用、也不宣稱證明一般微積分基本定理；不處理一般可積函數、Lebesgue 積分或面積理論；不改動 canonical MKO 或英文 edition。",
      "en": "Treats only one continuous polynomial within the MKO's scope, f(x)=x²−4 (integer coefficients, degree 2), and its signed Riemann integral over the finite interval [0,3]. Integrability of f is taken from MKO assumption a1 and is not proved here; what is proved is that, given integrability, the integral must equal −3. The general fundamental theorem of calculus is neither used nor claimed; general integrable functions, Lebesgue integration and area theory are not treated; the canonical MKO and its English edition are not changed."
    },
    "md_symbols": [
      [
        "f(x)=x^2-4",
        "被積多項式",
        "the integrand polynomial"
      ],
      [
        "I(f,a,b)",
        "f 從 a 到 b 的有向定積分",
        "the oriented definite integral of f from a to b"
      ],
      [
        "N",
        "等分份數，正整數",
        "number of equal pieces, a positive integer"
      ],
      [
        "\\Delta x=\\tfrac{3}{N}",
        "每段長度，也是網格",
        "length of each piece, which is also the mesh"
      ],
      [
        "x_i=\\tfrac{3i}{N}",
        "索引 i 的分點；0≤i<N 時也是第 i+1 段的左端標記",
        "the division point at index i; for 0≤i<N, also the left tag of piece i+1"
      ],
      [
        "L_N",
        "N 等分左端和",
        "the N-piece left sum"
      ],
      [
        "S(f,P,\\xi)",
        "標記分割 (P,ξ) 的 Riemann 和",
        "the Riemann sum of a tagged partition (P,ξ)"
      ]
    ],
    "md_error_controls": [
      {
        "title": "有限左端和不是積分 / A finite left sum is not the integral",
        "zh": "MKO 指出的常見誤解之一是把有限左端和當成積分。由 S09，誤差 |L_N+3|=(27N−9)/(2N²) 對每個 N≥1 都大於 0，所以沒有任何有限 N 的左端和等於 −3（例如 L_3=−7、L_50=−3.2682）；只有 N→∞ 的極限才等於積分。",
        "en": "One misconception the MKO names is taking a finite left sum for the integral. By S09 the error |L_N+3|=(27N−9)/(2N²) is positive for every N≥1, so no left sum with finite N equals −3 (for example L_3=−7 and L_50=−3.2682); only the limit as N→∞ equals the integral."
      },
      {
        "title": "反向端點要換號 / Reversed endpoints change the sign",
        "zh": "∫₃⁰ 與 ∫₀³ 的被積函數相同、方向相反，值為 3 而不是 −3。不能以「同一段區間」為由忽略方向，也不能一律取絕對值（MKO 假設 a2）。",
        "en": "∫₃⁰ and ∫₀³ have the same integrand and opposite orientation, so the value is 3, not −3. One may not ignore orientation on the grounds that it is 'the same interval', nor take absolute values by default (MKO assumption a2)."
      },
      {
        "title": "有號積分不是面積 / A signed integral is not an area",
        "zh": "f 在 [0,2) 為負、在 (2,3] 為正，兩段 −16/3 與 7/3 抵消成 −3；兩段絕對值之和 23/3 才是「不計正負」的量。把積分一律當非負面積，會把 −3 錯寫成 23/3 或 3。",
        "en": "f is negative on [0,2) and positive on (2,3]; the pieces −16/3 and 7/3 cancel to −3, while the unsigned quantity is the sum of absolute values 23/3. Treating every integral as a nonnegative area would wrongly give 23/3 or 3 instead of −3."
      },
      {
        "title": "矩核對不是一般 FTC / A moment check is not the general FTC",
        "zh": "S12 套用的是本 MKO 明列的多項式矩公式；它與 S11 一致只是兩條路互相核對。MKO 明說不把一般微積分基本定理當作已完成的前置，本稿也不宣稱證明了它。",
        "en": "S12 applies the polynomial moment formula listed in this MKO; its agreement with S11 is only a cross-check of two routes. The MKO says the general fundamental theorem of calculus is not treated as a completed prerequisite, and this guide does not claim to prove it."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "f 的可積性（G2）取自 MKO 假設 a1，本稿未由定義證明；S03 依賴它。",
        "en": "Integrability of f (G2) is taken from MKO assumption a1 and is not proved from the definition here; S03 depends on it."
      },
      {
        "zh-Hant": "矩公式（mko_moment）是 MKO 明列的規則；本稿只對 f(x)=x²−4 由定義獨立核對其值，未證明一般矩公式。",
        "en": "The moment formula (mko_moment) is a rule listed by the MKO; this guide checks its value independently from the definition for f(x)=x²−4 only and does not prove the general moment formula."
      },
      {
        "zh-Hant": "N4、N10 是 N=1 到 50 的有限核對，只支持這些點；一般 N 的結論依賴 S07、S09 的代數與 N3 的多項式恆等式。",
        "en": "N4 and N10 are finite checks for N=1 to 50 and support only those points; the conclusions for general N rest on the algebra of S07 and S09 and the polynomial identity N3."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "1358dfe43443e21f9926bd2ea2dd97a3b9034b0966a92b412d106d4ae7195ad4",
    "original_source_sha256": "1358dfe43443e21f9926bd2ea2dd97a3b9034b0966a92b412d106d4ae7195ad4",
    "owner_adopted": true,
    "bilingual_sync": "complete_source_bound",
    "formal_Lean": "NOT_RUN",
    "independent_AI_concurrence": "NOT_MEASURED",
    "issue_ids": [],
    "scope": "Owner-adopted natural-language teaching guide; not a formal proof, canonical MKO rewrite or evidence-status promotion."
  }
}