Let $f: [-1, 1] \to \mathbb R$ be a continuous function such $f(x) \ge 0$ for every $x \in [-1, 1]$ and $f(0)=1$ Show that the lower $\int_{-1}^1f>0$.
This is about being continuous, am I correct? But that's how far how I've gotten.
Continuous riemann integral
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riemann-integration
2 Answers
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Since $f \ge 0$ we have $\int_{-1}^1f(x) dx \ge 0$. From $f(0)=1$ and the continuity of $f$ we get some $a \in (0,1)$ such that $f(x) \ge 1/2$ for all $x \in (-a,a)$. It follows that
$\int_{-1}^1f(x) dx \ge \int_{-a}^a f(x) dx \ge \int_{-a}^a \frac{1}{2} dx =a>0$
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Another proof: Let $F(x)=\int_{-1}^xf(t) dt $ for $x \in [-1,1]$.
Suppose that $\int_{-1}^1f(x) dx = 0$. Since $f$ is continuous, $F$ is an antiderivative of $f$. Hence we have
- $0=F(1)-F(-1)$
and
- $F'=f \ge 0$.
Therefore $F$ is increasing, thus
$F(-1) \le F(x) \le F(1)$ for all $x \in [-1,1]$.
From 1. we see that $F$ is constant, hence $f(x)=0$ for all $x \in [-1,1]$, a contradiction.