Question Number 6610 by Temp last updated on 05/Jul/16
$$\mathrm{If}\:\mathrm{you}\:\mathrm{have}\:\mathrm{a}\:\mathrm{real}\:\mathrm{continuous}\:\mathrm{function}\:{f}\left({x}\right), \\ $$$$\mathrm{If}\:\mathrm{you}\:\mathrm{take}\:\mathrm{a}\:\mathrm{distance}\:\mathrm{between} \\ $$$${f}\left({x}\right)\:\mathrm{and}\:{f}\left({x}+\mathrm{1}\right),\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{average} \\ $$$$\mathrm{distance}\:\mathrm{within}\:{a}\leqslant{x}\leqslant{b}? \\ $$
Commented by Temp last updated on 05/Jul/16
$$\mathrm{let}\:\mathrm{the}\:\mathrm{distance}\:\mathrm{be}: \\ $$$${d}=\sqrt{\mid{f}\left({x}+\mathrm{1}\right)−{f}\left({x}\right)\mid^{\mathrm{2}} +\mid\left({x}+\mathrm{1}\right)−{x}\mid^{\mathrm{2}} } \\ $$$${d}=\sqrt{\mid{f}\left({x}+\mathrm{1}\right)−{f}\left({x}\right)\mid^{\mathrm{2}} +\mathrm{1}} \\ $$
Commented by Yozzii last updated on 05/Jul/16
$${d}_{{avg}} =\frac{\mathrm{1}}{{b}−{a}}\int_{{a}} ^{{b}} {d}\:{dx} \\ $$$${d}_{{avg}} =\frac{\mathrm{1}}{{b}−{a}}\int_{{a}} ^{{b}} \sqrt{\left({f}\left({x}+\mathrm{1}\right)−{f}\left({x}\right)\right)^{\mathrm{2}} +\mathrm{1}}\:{dx} \\ $$$$ \\ $$
Commented by Temp last updated on 05/Jul/16
$$\mathrm{I}\:\mathrm{thought}\:\mathrm{so},\:\mathrm{thank}\:\mathrm{you}! \\ $$$$ \\ $$
Commented by Temp last updated on 05/Jul/16
$$\mathrm{is}\:\mathrm{it}\:\left({f}\left({x}+\mathrm{1}\right)−{f}\left({x}\right)\right)^{\mathrm{2}} \\ $$$${or} \\ $$$$\mid{f}\left({x}+\mathrm{1}\right)−{f}\left({x}\right)\mid^{\mathrm{2}} \\ $$$$?? \\ $$
Commented by Yozzii last updated on 05/Jul/16
$${They}\:{are}\:{expressions}\:{equal}\:{in}\:{value} \\ $$$${so}\:{either}\:{notation}\:{is}\:{appropriate}.\: \\ $$$${I}\:{was}\:{taught}\:{that}\:{formula}\:{using}\:{parentheses}. \\ $$