Question Number 50007 by Joel578 last updated on 13/Dec/18 | ||
$$\mathrm{For}\:{a}\:<\:{x}\:<\:{b},\:\mathrm{find}\: \\ $$ $$\underset{{a}} {\overset{{b}} {\int}}\:\sqrt{{x}−{a}}\:.\:\sqrt{{b}−{x}}\:{dx} \\ $$ | ||
Commented byAbdo msup. last updated on 13/Dec/18 | ||
$${changement}\:\sqrt{{x}−{a}}={t}\:{give}\:{x}−{a}={t}^{\mathrm{2}} \:\Rightarrow \\ $$ $${I}\:=\int_{\mathrm{0}} ^{\sqrt{{b}−{a}}} {t}\sqrt{{b}−\left({a}+{t}^{\mathrm{2}} \right)}\mathrm{2}{t}\:{dt}\:=\mathrm{2}\int_{\mathrm{0}} ^{\sqrt{{b}−{a}}} {t}^{\mathrm{2}} \sqrt{{b}−{a}−{t}^{\mathrm{2}} }{dt} \\ $$ $$=_{{t}=\sqrt{{b}−{a}}{sinu}} \:\:\mathrm{2}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\left({b}−{a}\right){sin}^{\mathrm{2}} {u}\sqrt{{b}−{a}}{cosu}\:\sqrt{{b}−{a}}{cosu}\:{du} \\ $$ $$=\mathrm{2}\left({b}−{a}\right)^{\mathrm{2}} \:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{sin}^{\mathrm{2}} {u}\:{cos}^{\mathrm{2}} {u}\:{du} \\ $$ $$=\mathrm{2}\left({b}−{a}\right)^{\mathrm{2}} \:\frac{\mathrm{1}}{\mathrm{4}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{sin}^{\mathrm{2}} \left(\mathrm{2}{u}\right)\:{du} \\ $$ $$=\frac{\left({b}−{a}\right)^{\mathrm{2}} }{\mathrm{2}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{1}−{cos}\left(\mathrm{4}{u}\right)}{\mathrm{2}}\:{du} \\ $$ $$=\frac{\left({b}−{a}\right)^{\mathrm{2}} }{\mathrm{4}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\left(\mathrm{1}−{cos}\left(\mathrm{4}{u}\right)\right){du} \\ $$ $$=\frac{\pi}{\mathrm{8}}\left({b}−{a}\right)^{\mathrm{2}} \:−\frac{\left({b}−{a}\right)^{\mathrm{2}} }{\mathrm{16}}\left[{sin}\left(\mathrm{4}{u}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \\ $$ $$\bigstar{I}=\frac{\pi}{\mathrm{8}}\left({b}−{a}\right)^{\mathrm{2}} \:\bigstar \\ $$ | ||
Answered by ajfour last updated on 13/Dec/18 | ||
$${let}\:\:{x}−\:\frac{{a}+{b}}{\mathrm{2}}\:=\:{t} \\ $$ $$\int_{−\frac{{b}−{a}}{\mathrm{2}}} ^{\:\:\frac{{b}−{a}}{\mathrm{2}}} \sqrt{{t}+\frac{{b}−{a}}{\mathrm{2}}}\:\sqrt{\frac{{b}−{a}}{\mathrm{2}}−{t}}\:{dt} \\ $$ $${let}\:\:\:\:{c}\:=\:\frac{{b}−{a}}{\mathrm{2}} \\ $$ $$=\int_{−{c}} ^{\:\:{c}} \sqrt{{c}^{\mathrm{2}} −{t}^{\mathrm{2}} }\:{dt}\:=\:\mathrm{2}\int_{\mathrm{0}} ^{\:\:{c}} \sqrt{{c}^{\mathrm{2}} −{t}^{\mathrm{2}} }\:{dt} \\ $$ $$=\:\mathrm{2}\left[\frac{{t}}{\mathrm{2}}\sqrt{{c}^{\mathrm{2}} −{t}^{\mathrm{2}} }+\frac{{c}^{\mathrm{2}} }{\mathrm{2}}\mathrm{sin}^{−\mathrm{1}} \frac{{t}}{{c}}\right]_{\mathrm{0}} ^{{c}} \\ $$ $$=\:\:\frac{{c}^{\mathrm{2}} \pi}{\mathrm{2}}\:=\:\frac{\left({b}−{a}\right)^{\mathrm{2}} \pi}{\mathrm{8}}\:. \\ $$ | ||
Commented byMJS last updated on 13/Dec/18 | ||
$$\mathrm{grea}\:\mathrm{solutiont}! \\ $$ | ||
Commented byajfour last updated on 13/Dec/18 | ||
$${When}\:{we}\:{come}\:{across}\:{matters} \\ $$ $${that}\:{do}\:{not}\:{require}\:{our}\:{attention} \\ $$ $${at}\:{this}\:{moment},\:{although}\:{are} \\ $$ $${necessary}\:{for}\:{our}\:{conclusion}, \\ $$ $${then}\:{it}\:{is}\:{better}\:{to}\:{represent}\:{them} \\ $$ $${by}\:{the}\:{briefest}\:{possible}\:{symbols}. \\ $$ $$\:\:-\:{Rene}\:{Descartes}\:\left({Discourse}\:{on}\:{Method}\right) \\ $$ | ||
Commented byJoel578 last updated on 13/Dec/18 | ||
$${very}\:{well}\:{Sir} \\ $$ | ||