Answer
$\int_a^b f(x) dx \geq 0$
Work Step by Step
When the function $f(x)$ has minimum and maximum on $[a,b]$ then, we have
$min f(b-a) \leq \int_a^b f(x) dx \leq max f(b-a)$ ...(1)
Here, we have the function $f(x) \geq 0$ on $[a,b]$ then $ min f \geq 0$ and $ max f \geq 0$ on $[a,b]$
Then, we have $(b-a) \geq0; (b-a) min f \geq 0$ and $(b-a) max f
\geq 0$
From the equation (1), we get that both sides are positive.
Hence, the result has been verified $\int_a^b f(x) dx \geq 0$
