Answer
$\int_a^b f(x) dx \geq 0$
Work Step by Step
Consider the function $f(x) \geq 0$ defined on $[a,b]$ then $ min f \geq 0$ and $ max f \geq 0$ on $[a,b]$
When the function $f(x)$ has maximum and minimum values defined on $[a,b]$ then by Maximum-Minimum inequality , we have
$min f(b-a) \leq \int_a^b f(x) dx \leq max f(b-a)$ ....(1)
Then, we have $(b-a) \geq0$ and $(b-a) min f \geq 0$ and $(b-a) max f
\geq 0$
As we can see from the equation (1) that the both sides Maximum-Minimum inequality are positive.
Hence, the result has been verified $\int_a^b f(x) dx \geq 0$
