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Question Number 191435 by Matica last updated on 24/Apr/23 $$\:{It}\:{is}\:{given}\:{that}\:{I}_{\mathrm{1}} =\int{xg}\left[{x}\left(\mathrm{1}−{x}\right)\right]{dx}\:,\: \\ $$$${I}_{\mathrm{2}} =\:\int{g}\left[{x}\left(\mathrm{1}−{x}\right)\right]{dx}.\:{Find}\:\frac{{I}_{\mathrm{2}} }{{I}_{\mathrm{1}} }.\:{Thanks}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com

Question Number 125882 by harckinwunmy last updated on 14/Dec/20 $$\mathrm{suppose}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{set}\:\mathrm{of}\left\{\mathrm{3}>\mid\mathrm{c}+\mathrm{1}\mid\&\mid\mathrm{d}−\mathrm{1}\mid<\mathrm{10}\right\} \\ $$$$\mathrm{is}\:\mathrm{x}<\mathrm{c}+\mathrm{d}<\mathrm{y},\:\mathrm{then}\:\left(\mathrm{x},\mathrm{y}\right)=? \\ $$ Commented by talminator2856791 last updated on 15/Dec/20 $$\:\left(−\mathrm{12};\:\mathrm{11}\right)? \\ $$ Commented…

Question Number 125881 by aurpeyz last updated on 14/Dec/20 $${find}\:{the}\:{greatest}\:{coeeficient}\:{of} \\ $$$$\left(\mathrm{1}+\frac{{x}}{\mathrm{2}}\right)^{−\mathrm{2}} \\ $$$${i}\:{have}\:{applied}\:{Q}.\mathrm{125697}\:{but}\:{i}\:{got}\:{two} \\ $$$${negative}\:{values}\:{of}\:{n}\:{after}\:{i}\:{have} \\ $$$${made}\:{k}=\mathrm{2}{n}\:{for}\:{positive}\:{coefficient}. \\ $$$$ \\ $$$${I}\:{believe}\:{it}\:{should}\:{have}\:{a}\:{greatest} \\ $$$${coeeficient}\:{since}\:\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{k}} {decreases}\:{as}…

Question Number 191409 by null last updated on 23/Apr/23 Answered by senestro last updated on 24/Apr/23 $$−\frac{{x}}{\mathrm{6}\left(\mathrm{1}+{x}^{\mathrm{3}} \right)^{\mathrm{2}} }+\frac{{x}}{\mathrm{18}\left(\mathrm{1}+{x}^{\mathrm{3}} \right)}+\frac{\mathrm{1}}{\mathrm{9}}\mathrm{ln}\:\left(\frac{\left(\mathrm{1}+{x}\right)^{\mathrm{3}} }{\mathrm{1}+{x}^{\mathrm{3}} }\right)+\frac{\mathrm{2}}{\mathrm{3}\sqrt{\mathrm{3}}}{arctan}\left(\frac{\mathrm{2}{x}−\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)+{C} \\ $$ Terms…

Question Number 125867 by joki last updated on 14/Dec/20 $$\mathrm{partial}\:\mathrm{fraction}\:\mathrm{with}\:\mathrm{detail}\:\mathrm{step}\:\mathrm{by}\:\mathrm{step}\:\mathrm{from} \\ $$$$\frac{\mathrm{4x}^{\mathrm{3}} +\mathrm{x}^{\mathrm{2}} +\mathrm{1}}{\left(\mathrm{x}^{\mathrm{2}} −\mathrm{2}\right)^{\mathrm{3}} } \\ $$ Answered by mathmax by abdo last updated…