Wotzdorfg

2020-10-18

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

sweererlirumeX

Skilled2020-10-19Added 91 answers

Step 1

Given that f(x) has an absolute maximum of M and an absolute minimum of m

So, $m<f\left(x\right)<M$

Next, we integrate all sides between 0 and 2.

$m<f\left(x\right)<M$

${\int}_{0}^{2}mdx{\int}_{0}^{2}f\left(x\right)dx{\int}_{0}^{2}Mdx$

$m{\int}_{0}^{2}dx{\int}_{0}^{2}f\left(x\right)dxM{\int}_{0}^{2}dx$

$m{\left[x\right]}_{0}^{2}<{\int}_{0}^{2}f\left(x\right)dxM{\left[x\right]}_{0}^{2}$

$m(2-0)<{\int}_{0}^{2}f\left(x\right)dxM(2-0)$

$2m<{\int}_{0}^{2}f\left(x\right)dx2M$

Step 2

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

${\textstyle \phantom{\rule{1em}{0ex}}}\text{if}{\textstyle \phantom{\rule{1em}{0ex}}}m\le f\left(x\right)\le Mf{\textstyle \phantom{\rule{1em}{0ex}}}\text{or}{\textstyle \phantom{\rule{1em}{0ex}}}a\le x\le b,then$

$m(b-a)\le {\int}_{a}^{b}f\left(x\right)dx\le M(b-a)$

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