Blog: `f(x) = (2x)/(1 + x^2)^2, [0, 2]` Find the average value of the function on the given interval.

Wednesday, January 18, 2017

Given $\displaystyle{f{{\left({x}\right)}}}=\frac{{{2}{x}}}{{{\left({1}+{{x}}^{{2}}\right)}}^{{2}}},{\left[{0},{2}\right]}$

Average Value Formula= $\displaystyle\frac{{1}}{{{b}-{a}}}{\int_{{a}}^{{b}}}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}$

$\displaystyle{f}_{{{a}{v}{e}}}=\frac{{1}}{{{2}-{0}}}{\int_{{0}}^{{2}}}\frac{{{2}{x}}}{{{\left({1}+{{x}}^{{2}}\right)}}^{{2}}}{\left.{d}{x}\right.}$

$\displaystyle={\left(\frac{{1}}{{2}}\right)}\cdot{2}{\int_{{0}}^{{2}}}\frac{{x}}{{{\left({1}+{{x}}^{{2}}\right)}}^{{2}}}{\left.{d}{x}\right.}$

$\displaystyle={\int_{{0}}^{{2}}}\frac{{x}}{{{\left({1}+{{x}}^{{2}}\right)}}^{{2}}}{\left.{d}{x}\right.}$

Integrate using the u-subsitution method.

Let $\displaystyle{u}={1}+{{x}}^{{2}}$

$\displaystyle\frac{{{d}{u}}}{{\left.{d}{x}}}={2}{x}$

$\displaystyle{\left.{d}{x}\right.}=\frac{{{d}{u}}}{{{2}{x}}}$

$\displaystyle={\int_{{0}}^{{2}}}\frac{{x}}{{{u}}^{{2}}}\cdot\frac{{{d}{u}}}{{{2}{x}}}$

$\displaystyle={\left(\frac{{1}}{{2}}\right)}{\int_{{0}}^{{2}}}{\left(\frac{{1}}{{{u}}^{{2}}}\right)}{d}{u}$

$\displaystyle={\left(\frac{{1}}{{2}}\right)}{\int_{{0}}^{{2}}}{\left({{u}}^{{-{{2}}}}\right)}{d}{u}$

$\displaystyle={\left(\frac{{1}}{{2}}\right)}\frac{{{{u}}^{{-{{1}}}}}}{{-{1}}}{{\mid}_{{0}}^{{2}}}$

$\displaystyle={\left(-\frac{{1}}{{2}}\right)}{{\left[\frac{{1}}{{u}}\right]}_{{0}}^{{2}}}$

$\displaystyle={\left(-\frac{{1}}{{2}}\right)}{{\left[\frac{{1}}{{{1}+{{x}}^{{2}}}}\right]}_{{0}}^{{2}}}$

$\displaystyle={\left(-\frac{{1}}{{2}}\right)}{\left[\frac{{1}}{{{1}+{{\left({2}\right)}}^{{2}}}}-\frac{{1}}{{{1}-{{0}}^{{2}}}}\right]}$

$\displaystyle={\left(-\frac{{1}}{{2}}\right)}{\left[\frac{{1}}{{5}}-{1}\right]}$

$\displaystyle={\left(-\frac{{1}}{{2}}\right)}{\left(-\frac{{4}}{{5}}\right)}$

$\displaystyle=\frac{{2}}{{5}}$

The average value is 2/5.