§5.3.2, proof of Theorem 5.3, p. 321 — ∇²f(x_k)(x_{k+1} − x*) = ∫₀¹ [∇²f(x_k) − ∇²f(x* + s(x_k − x*))](x_k − x*) ds

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§5.3.2, proof of Theorem 5.3, p. 321 — ∇²f(x_k)(x_{k+1} − x*) = ∫₀¹ [∇²f(x_k) − ∇²f(x* + s(x_k − x*))](x_k − x*) ds · Prove2Me