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Soit \(f : {[0,1]} \to \R\) continue positive. On pose \(A = \int_{t=0}^1 f(t)\,d t\).
Montrer que \(\sqrt{1+A^2} \le \int_{t=0}^1 \sqrt{1+f^2(t)}\,d t \le 1+A\).
Soit \(f : {[0,1]} \to \R\) continue positive. On pose \(A = \int_{t=0}^1 f(t)\,d t\).
Montrer que \(\sqrt{1+A^2} \le \int_{t=0}^1 \sqrt{1+f^2(t)}\,d t \le 1+A\).