Question Number 24411 by A1B1C1D1 last updated on 17/Nov/17 Answered by mrW1 last updated on 17/Nov/17 $$\int_{\frac{\pi}{\mathrm{2}}} ^{\:\mathrm{0}} \int_{\mathrm{2}} ^{\:\mathrm{0}} {r}^{\mathrm{4}} \mathrm{cos}\:\left(\mathrm{2}\theta\right)\:{dr}\:{d}\theta \\ $$$$=\int_{\frac{\pi}{\mathrm{2}}} ^{\:\mathrm{0}}…
Question Number 89918 by jagoll last updated on 20/Apr/20 $$\int\:\frac{\mathrm{x}\:\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{x}\right)}{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{3}/\mathrm{2}} }\:\mathrm{dx}\: \\ $$ Commented by john santu last updated on 20/Apr/20 $$\left[\:{u}\:=\:\mathrm{tan}^{−\mathrm{1}} \left({x}\right)\:\Rightarrow{du}\:=\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}}…
Question Number 24363 by gopikrishnan005@gmail.com last updated on 16/Nov/17 $$\int\left({ae}\right)^{{x}} {dx} \\ $$ Answered by ajfour last updated on 16/Nov/17 $$=\frac{\left({ae}\right)^{{x}} }{\mathrm{1}+\mathrm{ln}\:{a}}+{C}\:. \\ $$ Commented…
Question Number 89891 by mathmax by abdo last updated on 19/Apr/20 $${calculate}\:\int_{\frac{\mathrm{1}}{{e}}} ^{\mathrm{1}} {ln}\left({x}\right){ln}\left(\mathrm{1}+{x}\right){dx} \\ $$ Commented by mathmax by abdo last updated on 20/Apr/20…
Question Number 89874 by M±th+et£s last updated on 19/Apr/20 $$\int_{\frac{\mathrm{1}}{\mathrm{3}}} ^{\frac{\mathrm{2}}{\mathrm{3}}} \:\frac{{ln}\left(\mathrm{1}+\sqrt{{x}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{3}}}\right)}{{x}\sqrt{{x}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{3}}}}{dx} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 89867 by M±th+et£s last updated on 19/Apr/20 $${find}\:{the}\:{integration} \\ $$$$\int{ln}\lfloor{x}\rfloor\:{dx}\:\:\:\:\:\:;\:{x}>\mathrm{2} \\ $$ Commented by ~blr237~ last updated on 20/Apr/20 Answered by mr W…
Question Number 89846 by M±th+et£s last updated on 19/Apr/20 $$\int_{−\mathrm{1}} ^{\mathrm{1}} \:{x}\:{cosh}\left({x}\right)\:{ln}\left(\mathrm{1}+{e}^{{x}} \right)\:{dx} \\ $$ Commented by M±th+et£s last updated on 19/Apr/20 $${thank}\:{you}\:{verry}\:{much} \\ $$…
Question Number 89749 by cindiaulia last updated on 19/Apr/20 Commented by mr W last updated on 19/Apr/20 Commented by mr W last updated on 19/Apr/20…
Question Number 89736 by cindiaulia last updated on 19/Apr/20 Commented by jagoll last updated on 19/Apr/20 $$\mathrm{vol}\:=\:\pi\:\left[\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\left(\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{2}} \:\mathrm{dx}\:+\:\underset{\mathrm{1}} {\overset{\mathrm{4}} {\int}}\:\left(\sqrt{\mathrm{x}}\right)^{\mathrm{2}} \:\mathrm{dx}\:\right] \\…
Question Number 89741 by cindiaulia last updated on 19/Apr/20 Commented by jagoll last updated on 19/Apr/20 $$\mathrm{area}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\mid\begin{vmatrix}{\mathrm{0}\:\:\:\:\:\mathrm{0}}\\{\mathrm{1}\:\:\:\:\:\mathrm{1}}\end{vmatrix}+\:\begin{vmatrix}{\mathrm{1}\:\:\:\:\:\:\mathrm{1}}\\{\frac{\mathrm{2}}{\mathrm{5}}\:\:\frac{\mathrm{8}}{\mathrm{5}}}\end{vmatrix}+\:\begin{vmatrix}{\frac{\mathrm{2}}{\mathrm{5}}\:\:\:\:\frac{\mathrm{8}}{\mathrm{5}}}\\{\mathrm{0}\:\:\:\:\:\:\:\mathrm{0}}\end{vmatrix}\mid \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\:\mid\:\mathrm{0}+\frac{\mathrm{6}}{\mathrm{5}}+\mathrm{0}\:\mid\:=\:\frac{\mathrm{3}}{\mathrm{5}} \\ $$ Terms of Service Privacy…