Chapter 6.3, Problem 10E

To determine

To show: If f is integrable then $f(x)\le 0\text{on }[a,b]\Rightarrow {{\displaystyle \int }}_a^{b}f(x)dx\le 0$.

Explanation of Solution

Giveninformation :

Function f is integrable

Calculation :

Since on $[a,b]$

  $f(x)\le 0$

So, if min f is the maximum value of f on $\left[a,b\right]$

  $\max f(x)\le 0$

Now, multiply both sides of inequality by $(b-a)$

  $(b-a)\max f(x)\le 0$

Since,

  ${{\displaystyle \int }}_a^{b}f(x)dx\le \max f(b-a)$

So,

  ${{\displaystyle \int }}_a^{b}f(x)dx\le 0$