Question Number 109427 by mathdave last updated on 23/Aug/20 Answered by 1549442205PVT last updated on 23/Aug/20 $$\begin{vmatrix}{\mathrm{x}}&{\mathrm{0}}&{}&{\mathrm{1}}&{}&{\mathrm{2}}&{}&{\mathrm{3}}&{}&{\mathrm{4}}\\{\mid\mathrm{x}−\mathrm{1}\mid}&{}&{\mathrm{1}−\mathrm{x}}&{\mathrm{0}}&{\mathrm{x}−\mathrm{1}}&{}&{\mathrm{x}−\mathrm{1}}&{}&{\mathrm{x}−\mathrm{1}}&{}\\{\mid\mathrm{x}−\mathrm{2}\mid}&{}&{\mathrm{2}−\mathrm{x}}&{}&{\mathrm{2}−\mathrm{x}}&{\mathrm{9}}&{\mathrm{x}−\mathrm{2}}&{}&{\mathrm{x}−\mathrm{2}}&{}\\{\mid\mathrm{x}−\mathrm{3}\mid}&{}&{\mathrm{3}−\mathrm{x}}&{}&{\mathrm{3}−\mathrm{x}}&{}&{\mathrm{3}−\mathrm{x}}&{}&{\mathrm{x}−\mathrm{3}}&{}\\{\mathrm{nom}}&{}&{\mathrm{3}−\mathrm{2x}}&{}&{\mathrm{1}}&{}&{\mathrm{2x}−\mathrm{3}}&{}&{\mathrm{2x}−\mathrm{3}}&{}\\{\mathrm{deno}}&{}&{\mathrm{8}−\mathrm{4x}}&{}&{\mathrm{6}−\mathrm{2x}}&{}&{−\mathrm{2}+\mathrm{2x}}&{}&{\mathrm{4x}−\mathrm{8}}&{}\\{\mathrm{f}\left(\mathrm{x}\right)}&{}&{\frac{\mathrm{2x}−\mathrm{3}}{\mathrm{4x}−\mathrm{8}}}&{}&{\frac{\mathrm{1}}{\mathrm{6}−\mathrm{2x}}}&{}&{\frac{\mathrm{2x}−\mathrm{3}}{\mathrm{2x}−\mathrm{2}}}&{}&{\frac{\mathrm{2x}−\mathrm{3}}{\mathrm{4x}−\mathrm{8}}}&{}\end{vmatrix} \\ $$$$\mathrm{From}\:\mathrm{above}\:\mathrm{tablet}\:\mathrm{we}\:\mathrm{have} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{4}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2x}−\mathrm{3}}{\mathrm{4x}−\mathrm{8}}\mathrm{dx}+\int_{\mathrm{1}}…
Question Number 109403 by abony1303 last updated on 23/Aug/20 $$\int\:\frac{\mathrm{dx}}{\:\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{a}^{\mathrm{2}} }}=\mathrm{ln}\mid\mathrm{x}+\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{a}^{\mathrm{2}} }\mid+{C}\:\:\:\:\mathrm{Proof}? \\ $$ Answered by bobhans last updated on 23/Aug/20 $${let}\:{x}\:=\:{a}\:\mathrm{tan}\:{s}\:\Rightarrow\:{dx}\:=\:{a}\:\mathrm{sec}\:^{\mathrm{2}} {s}\:{ds}\:…
Question Number 109400 by mohammad17 last updated on 23/Aug/20 Commented by mohammad17 last updated on 23/Aug/20 $${help}\:{me}\:{sir}? \\ $$ Commented by bobhans last updated on…
Question Number 109396 by ZiYangLee last updated on 23/Aug/20 $$\mathrm{If}\:{z}^{\frac{\mathrm{1}}{\mathrm{2}}} ={x}^{\frac{\mathrm{1}}{\mathrm{2}}} +{y}^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\left({x}+{y}−{z}\right)^{\mathrm{2}} =\mathrm{4}{xy} \\ $$ Answered by bemath last updated on 23/Aug/20 $${z}\:=\:{x}+{y}+\mathrm{2}\sqrt{{xy}}\:…
Question Number 43851 by jasno91 last updated on 16/Sep/18 Answered by tanmay.chaudhury50@gmail.com last updated on 16/Sep/18 $${l}=\mathrm{6}\:\:\:{b}=\frac{\mathrm{2}}{\mathrm{3}}×\mathrm{6}=\mathrm{4} \\ $$$${p}=\mathrm{2}\left({l}+{b}\right)=\mathrm{2}×\mathrm{10}=\mathrm{20}\:{cm} \\ $$$$ \\ $$$$ \\ $$$$…
Question Number 109373 by mathdave last updated on 23/Aug/20 Commented by mnjuly1970 last updated on 23/Aug/20 $${murray}\:{spiegel}.\:{advanced} \\ $$$${calculus}. \\ $$ Commented by PRITHWISH SEN…
Question Number 174902 by daus last updated on 14/Aug/22 Answered by Ar Brandon last updated on 14/Aug/22 $$\int_{\mathrm{0}} ^{\mathrm{4}} \frac{\mathrm{1}}{\left(\mathrm{1}+\sqrt{{x}}\right)^{\mathrm{2}} }{dx}\:,\:{x}={t}^{\mathrm{2}} \:\Rightarrow{dx}=\mathrm{2}{tdt} \\ $$$$\:\:\:\:=\int_{\mathrm{0}} ^{\mathrm{2}}…
Question Number 43822 by peter frank last updated on 15/Sep/18 Answered by Joel578 last updated on 17/Sep/18 $$\left({b}\right) \\ $$$$\mathrm{4cosh}\:{x}\:+\:\mathrm{3cosh}\:{y}\:=\:\mathrm{10}\:\:\:\:…\left({i}\right) \\ $$$$\mathrm{4sinh}\:{x}\:+\:\mathrm{3sinh}\:{y}\:=\:\mathrm{7}\:\:\:\:\:\:\:\:…\left({ii}\right) \\ $$$$ \\…
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Question Number 174884 by daus last updated on 13/Aug/22 Answered by aleks041103 last updated on 13/Aug/22 $$\left({log}_{{b}} {a}\right)\left({log}_{{a}} {b}\right)={log}_{{a}} \left({b}^{{log}_{{b}} {a}} \right)={log}_{{a}} {a}=\mathrm{1} \\ $$$$\Rightarrow{log}_{{x}}…