§5.3.2, proof of Theorem 5.3, p. 321 — ∫₀¹ ∇²f(x + sh) h ds = ∇f(x + h) − ∇f(x)

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§5.3.2, proof of Theorem 5.3, p. 321 — ∫₀¹ ∇²f(x + sh) h ds = ∇f(x + h) − ∇f(x) · Prove2Me