# what-is-the-mean-value-of-1-1-4x-2-for-0-x-1-2-

Question Number 76721 by Rio Michael last updated on 29/Dec/19
$$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{mean}\:\mathrm{value}\:\mathrm{of}\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{4x}^{\mathrm{2}\:} }\:\:\mathrm{for}\:\:\mathrm{0}\leqslant\mathrm{x}\leqslant\frac{\mathrm{1}}{\mathrm{2}} \\$$
Answered by MJS last updated on 30/Dec/19
$$\mathrm{mean}\:\left({f}\left({x}\right)\right)\:\mathrm{with}\:{a}\leqslant{x}\leqslant{b} \\$$$$\frac{\mathrm{1}}{{b}−{a}}\underset{{a}} {\overset{{b}} {\int}}{f}\left({x}\right){dx} \\$$$$\mathrm{we}'\mathrm{re}\:\mathrm{calculating}\:\mathrm{the}\:\mathrm{area}\:\mathrm{first}\:\mathrm{and}\:\mathrm{then} \\$$$$\mathrm{the}\:\mathrm{height}\:\mathrm{of}\:\mathrm{a}\:\mathrm{rectangle}\:\mathrm{of}\:\mathrm{the}\:\mathrm{width}\:\mathrm{of} \\$$$$\mathrm{our}\:\mathrm{interval} \\$$$$\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{0}}\underset{\mathrm{0}} {\overset{\frac{\mathrm{1}}{\mathrm{2}}} {\int}}\frac{{dx}}{\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} }=\mathrm{2}\underset{\mathrm{0}} {\overset{\frac{\mathrm{1}}{\mathrm{2}}} {\int}}\frac{{dx}}{\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} }= \\$$$$\:\:\:\:\:\left[{t}=\mathrm{2}{x}\:\rightarrow\:{dx}=\frac{\mathrm{1}}{\mathrm{2}}{dt}\right] \\$$$$=\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{1}}=\left[\mathrm{arctan}\:{t}\right]_{\mathrm{0}} ^{\mathrm{1}} =\frac{\pi}{\mathrm{4}} \\$$
Answered by john santu last updated on 30/Dec/19
$$=\:\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{0}}\:\int\underset{\mathrm{0}} {\overset{\frac{\mathrm{1}}{\mathrm{2}}} {\:}}\:\frac{{dx}}{\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} } \\$$$$=\:\mathrm{2}\:\int\underset{\mathrm{0}} {\overset{\frac{\mathrm{1}}{\mathrm{2}}} {\:}}\:\frac{{dx}}{\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} }\:=\:\int\underset{\mathrm{0}} {\overset{\frac{\mathrm{1}}{\mathrm{2}}} {\:}}\frac{{d}\left(\mathrm{2}{x}\right)}{\mathrm{1}+\left(\mathrm{2}{x}\right)^{\mathrm{2}} } \\$$$$=\:{arc}\mathrm{tan}\:\left(\mathrm{2}{x}\right)\:\mid\underset{\mathrm{0}} {\overset{\frac{\mathrm{1}}{\mathrm{2}}} {\:}}=\frac{\pi}{\mathrm{4}}\:\blacksquare \\$$
Commented by Rio Michael last updated on 30/Dec/19
$$\mathrm{thanks}\: \\$$