Question Number 208908 by alcohol last updated on 26/Jun/24 $${I}\left({a},{b}\right)\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} {t}^{{a}} \left(\mathrm{1}−{t}\right)^{{b}} {dt} \\ $$$${Note}\::\:{I}\left({a},{b}\right)\:=\:\frac{{a}}{{b}+\mathrm{1}}{I}\left({a}−\mathrm{1},{b}+\mathrm{1}\right) \\ $$$${show}\:{that} \\ $$$$\bullet\:{I}\left({a}+\mathrm{1},\:{b}\right)\:+\:{I}\left({a},{b}+\mathrm{1}\right)\:=\:{I}\left({a},\:{b}\right) \\ $$$$\bullet\:{find}\:{B}\left({a}+\mathrm{1},\:{b}+\mathrm{1}\right)\:{interms}\:{of}\:{B}\left({a},{b}\right) \\ $$$$\bullet\:{use}\:{I}\left({a},{b}\right)\:=\:{I}\left({b},{a}\right)\:{and}\:{deduce}\:{I}\left(\frac{\mathrm{5}}{\mathrm{2}},\frac{\mathrm{3}}{\mathrm{2}}\right) \\…
Question Number 208892 by hardmath last updated on 26/Jun/24 $$\mathrm{Find}: \\ $$$$\sqrt{−\mathrm{16}}\:\:\centerdot\:\:\sqrt{−\mathrm{9}}\:\:=\:\:? \\ $$ Commented by Adeyemi889 last updated on 26/Jun/24 $$ \\ $$$$\sqrt{−\mathrm{16}}\:=\:\sqrt{\:\left(\mathrm{16}\right)\left(−\mathrm{1}\right)}\:=\sqrt{−\mathrm{1}}\:×\sqrt{\mathrm{16}\:} \\…
Question Number 208876 by Tawa11 last updated on 26/Jun/24 The 𝚌a𝚕𝚎𝚗𝚍𝚊𝚛 𝚘𝚏 𝚝𝚑𝚎 𝚢𝚎𝚊𝚛 2024 𝚒𝚜 𝚝𝚑𝚎 𝚜𝚊𝚖𝚎 𝚏𝚘𝚛 𝙰.2044 𝙱.2032 𝙲.2040 𝙳.2036 Commented by…
Question Number 208909 by lepuissantcedricjunior last updated on 26/Jun/24 $$\:\:\:\:\:\boldsymbol{{soit}}\:\boldsymbol{{la}}\:\boldsymbol{{fonction}}\:\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)=\boldsymbol{{x}}^{\mathrm{3}} +\boldsymbol{{x}}\:\:\boldsymbol{{definie}} \\ $$$$\boldsymbol{{sur}}\:\mathbb{R}\:\boldsymbol{{on}}\:\boldsymbol{{note}}\:\boldsymbol{{g}}\left(\boldsymbol{{x}}\right)=\boldsymbol{{f}}^{−\mathrm{1}} \left(\boldsymbol{{x}}\right) \\ $$$$\boldsymbol{{alors}}\:\boldsymbol{{que}}\:\:\boldsymbol{{la}}\:\boldsymbol{{primitive}}\:\boldsymbol{{G}}\left(\boldsymbol{{x}}\right)=\int_{\mathrm{0}} ^{\boldsymbol{{x}}} \boldsymbol{{g}}\left(\boldsymbol{{t}}\right)\boldsymbol{{dt}} \\ $$ Commented by Ghisom last updated…
Question Number 208872 by Ismoiljon_008 last updated on 26/Jun/24 $$ \\ $$$$\:\:\:{Find}\:{the}\:{side}\:{of}\:{a}\:{triangle}\:{if}\:{the}\:{distances} \\ $$$$\:\:\:{from}\:{an}\:{arbitrary}\:{point}\:{inside}\:{a}\:{regular}\:{triangle}\: \\ $$$$\:\:\:{to}\:{its}\:{vertices}\:{are}\:{m},\:{n}\:{and}\:{k}. \\ $$$$\:\:{Help}\:{please} \\ $$ Answered by mr W last…
Question Number 208891 by efronzo1 last updated on 26/Jun/24 $$\:\:\:\underline{\kappa} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 208900 by KradecaNaLukcheta last updated on 26/Jun/24 $${Does}\:{anyone}\:{know}\:{of}\:{an}\:{intuition} \\ $$$${behind}\:{the}\:{integral}\:{form}\:{of}\:{the} \\ $$$${remainder}\:{in}\:{Taylor}'{s}\:{theorem}? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 208871 by Shrodinger last updated on 26/Jun/24 $${L}=\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\frac{\mathrm{4}−\mathrm{3}{x}}{\mathrm{4}+\mathrm{5}{x}}}{dx} \\ $$ Answered by Sutrisno last updated on 26/Jun/24 $${misal} \\ $$$$\sqrt{\frac{\mathrm{4}−\mathrm{3}{x}}{\mathrm{4}+\mathrm{5}{x}}}={p}\rightarrow{x}=\frac{−\mathrm{4}{p}^{\mathrm{2}} +\mathrm{4}}{\mathrm{5}{p}^{\mathrm{2}}…
Question Number 208896 by efronzo1 last updated on 26/Jun/24 Answered by MM42 last updated on 27/Jun/24 $${s}_{\mathrm{1}} =\mathrm{32}−\int_{\mathrm{0}} ^{\mathrm{4}} \sqrt{\mathrm{64}−{x}^{\mathrm{2}} }{dx}\:\:\:\:\:;\:\:{x}=\mathrm{8}{sin}\theta \\ $$$$\Rightarrow=\mathrm{32}−\mathrm{64}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \:{cos}^{\mathrm{2}}…
Question Number 208880 by efronzo1 last updated on 26/Jun/24 Answered by Frix last updated on 26/Jun/24 $${a},\:{b},\:{c}\:>\mathrm{0}\:\Rightarrow\:\sqrt[{\mathrm{3}}]{{a}+\sqrt[{\mathrm{3}}]{\mathrm{2}}+\sqrt[{\mathrm{3}}]{\mathrm{4}}}+\sqrt[{\mathrm{3}}]{{b}}+\sqrt[{\mathrm{3}}]{{c}}>\mathrm{0} \\ $$$$\mathrm{The}\:\mathrm{question}\:\mathrm{has}\:\mathrm{no}\:\mathrm{solution}. \\ $$ Terms of Service Privacy…