Suppose that \(f\) is continuous and \(f(x)\geq 0\) on \([0,1]\). If \(f(0)>0\), prove that \(\int_0^1 f(x)dx>0\).
Solution
1. Since \(f(0) > 0\) and \(f\) is continuous, there is \(\delta \in (0,1)\) with \(f(x) > f(0)/2\) for all \(x \in [0,\delta]\).
continuity at \(0\) with \(\eps = f(0)/2 > 0\) gives \(|f(x) - f(0)| < f(0)/2\) for \(|x| < \delta\), hence \(f(x) > f(0) - f(0)/2 = f(0)/2\).
2. \(\int_0^1 f \ge \int_0^\delta f \ge \delta \cdot f(0)/2 > 0\).
\(f \ge 0\) on \([0,1]\) gives \(\int_0^1 f \ge \int_0^\delta f\); and on \([0,\delta]\), \(f \ge f(0)/2\) by 1, so \(\int_0^\delta f \ge \delta f(0)/2 > 0\).
3. Q.E.D.