## Thomas' Calculus 13th Edition

$\int_a^b f(x) dx \leq 0$
When the function $f(x)$ has minimum and maximum values defined on $[a,b]$ then the maximum-minimum inequality tells that $min f(b-a) \leq \int_a^b f(x) dx \leq max f(b-a)$ Here, we have the function $f(x) \leq 0$ on $[a,b]$ then $min f \leq 0$ and $max f \leq 0$ on $[a,b]$ This implies that $(b-a) \geq0$ and $(b-a) min f \leq 0$ Also, $(b-a) max f \leq 0$ Here, the both sides of the inequality $min f(b-a) \leq \int_a^b f(x) dx \leq max f(b-a)$ are positive. Hence, the result is proved that $\int_a^b f(x) dx \leq 0$