Suppose f has absolute minimum value m and absolute maximum value M. Between what two values must int 0^2 f(x)dx lie? Which property of integrals allows you to make your conclusion?

Wotzdorfg 2020-10-18 Answered
Suppose f has absolute minimum value m and absolute maximum value M. Between what two values must \(\displaystyle\int{0}^{{2}}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}\) lie? Which property of integrals allows you to make your conclusion?

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Expert Answer

sweererlirumeX
Answered 2020-10-19 Author has 21793 answers

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

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