Chapter 6.3, Problem 9E

To determine

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

Explanation of Solution

Given information :

Function f is integrable

Calculation :

Since on $[a,b]$

  $f(x)\ge 0$

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

  $\min f(x)\ge 0$

Now, multiply both sides of inequality by $\left[a,b\right]$

  $(b-a)\min f(x)\ge 0$

Since,

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

So,

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