Step 1

Given that the absolute minimum value of f(x) is m and absolute maximum value is M

So, \(\displaystyle{m}{<}{f{{\left({x}\right)}}}{<}{M}\)

Then we integrate all sides from 0 to 2

\(\displaystyle{m}{<}{f{{\left({x}\right)}}}{<}{M}\)

\(\displaystyle{\int_{{0}}^{{2}}}{m}{\left.{d}{x}\right.}{<}{\int_{{0}}^{{2}}}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}{<}{\int_{{0}}^{{2}}}{M}{\left.{d}{x}\right.}\)

\(\displaystyle{m}{\int_{{0}}^{{2}}}{\left.{d}{x}\right.}{<}{\int_{{0}}^{{2}}}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}{<}{M}{\int_{{0}}^{{2}}}{\left.{d}{x}\right.}\)

\(\displaystyle{m}{{\left[{x}\right]}_{{0}}^{{2}}}{<}{\int_{{0}}^{{2}}}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}{<}{M}{{\left[{x}\right]}_{{0}}^{{2}}}\)

\(\displaystyle{m}{\left({2}-{0}\right)}{<}{\int_{{0}}^{{2}}}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}{<}{M}{\left({2}-{0}\right)}\)

\(\displaystyle{2}{m}{<}{\int_{{0}}^{{2}}}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}{<}{2}{M}\)

Step 2

Using the extreme value theorem and comparison property of integrals we made our conclusion comparison property of integrals:

\(\displaystyle{\quad\text{if}\quad}{m}\le{f{{\left({x}\right)}}}\le{M}{f}{\quad\text{or}\quad}{a}\le{x}\le{b},{t}{h}{e}{n}\)

\(\displaystyle{m}{\left({b}-{a}\right)}\le{\int_{{a}}^{{b}}}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}\le{M}{\left({b}-{a}\right)}\)