Question Number 57319 by turbo msup by abdo last updated on 02/Apr/19 $${calculate}\:\int\int_{{D}} \:{e}^{{x}−{y}} \:{dxdy} \\ $$$${with}\:{D}=\left\{\left({x},{y}\right)\in{R}^{\mathrm{2}} \:/\mid{x}\mid<\mathrm{1}\:{and}\:\mathrm{0}\leqslant{y}\leqslant\mathrm{1}\right\} \\ $$ Commented by maxmathsup by imad…
Question Number 188384 by universe last updated on 28/Feb/23 $$\:\:\: \\ $$$$\:\:\:\mathrm{if}\:\:\:\:\int_{\mathrm{0}} ^{\infty} {e}^{−{ax}} {dx}\:\:=\:\:\frac{\mathrm{1}}{{a}} \\ $$$$\:\:\:\:\:\:\:\:{show}\:{that}\:\:\int_{\mathrm{0}} ^{\infty} {e}^{−{ax}} \:{x}^{{n}} {dx}\:\:=\:\:\frac{{n}!}{{a}^{{n}+\mathrm{1}} } \\ $$ Answered…
Question Number 188381 by mnjuly1970 last updated on 28/Feb/23 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{Advanced}\:\:\mathrm{calculus} \\ $$$$ \\ $$$$\:\:\:\:\:\mathrm{Find}\:\:\mathrm{the}\:\:\mathrm{value}\:\:\mathrm{of}\:\mathrm{the}\:\mathrm{following}\:\mathrm{series}. \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\Omega\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\:\left(−\mathrm{1}\right)^{\:{n}} \:\zeta\:\left(\:{n}\:\right)}{{n}.\:\mathrm{2}^{\:{n}} }\:=\:? \\…
Question Number 188380 by Shlock last updated on 28/Feb/23 Answered by SEKRET last updated on 28/Feb/23 $$\:\:\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{t}}\right)=\int_{\mathrm{0}} ^{\:\boldsymbol{\pi}} \frac{\boldsymbol{\mathrm{ln}}\left(\mathrm{1}+\boldsymbol{\mathrm{t}}\centerdot\boldsymbol{\mathrm{sin}}\left(\boldsymbol{\mathrm{a}}\right)\centerdot\boldsymbol{\mathrm{cos}}\left(\boldsymbol{\mathrm{x}}\right)\right)}{\boldsymbol{\mathrm{cos}}\left(\boldsymbol{\mathrm{x}}\right)}\boldsymbol{\mathrm{dx}} \\ $$$$\:\boldsymbol{\mathrm{f}}\:'\:\left(\boldsymbol{\mathrm{t}}\right)=\:\int_{\mathrm{0}} ^{\:\boldsymbol{\pi}} \:\frac{\boldsymbol{\mathrm{sin}}\left(\boldsymbol{\mathrm{a}}\right)\centerdot\boldsymbol{\mathrm{cos}}\left(\boldsymbol{\mathrm{x}}\right)}{\boldsymbol{\mathrm{cos}}\left(\boldsymbol{\mathrm{x}}\right)\centerdot\left(\mathrm{1}+\boldsymbol{\mathrm{t}}\centerdot\boldsymbol{\mathrm{sin}}\left(\boldsymbol{\mathrm{a}}\right)\centerdot\boldsymbol{\mathrm{cos}}\left(\boldsymbol{\mathrm{x}}\right)\right)}\boldsymbol{\mathrm{dx}} \\ $$$$\:\boldsymbol{\mathrm{f}}\:'\:\left(\boldsymbol{\mathrm{t}}\right)=\:\boldsymbol{\mathrm{sin}}\left(\boldsymbol{\mathrm{a}}\right)\centerdot\int_{\mathrm{0}}…
Question Number 122838 by bemath last updated on 20/Nov/20 $$\:\:\int\:\frac{{dx}}{\:\sqrt[{\mathrm{4}}]{\left({x}−\mathrm{1}\right)^{\mathrm{3}} \left({x}+\mathrm{2}\right)^{\mathrm{5}} }}\:? \\ $$ Answered by liberty last updated on 20/Nov/20 $$\:{because}\:\sqrt[{\mathrm{4}}]{\left({x}−\mathrm{1}\right)^{\mathrm{3}} \left({x}+\mathrm{2}\right)^{\mathrm{5}} }\:=\:\left({x}−\mathrm{1}\right)\left({x}+\mathrm{2}\right)\:\sqrt[{\mathrm{4}}]{\frac{{x}+\mathrm{2}}{{x}−\mathrm{1}}} \\…
Question Number 122828 by bemath last updated on 20/Nov/20 Answered by bobhans last updated on 20/Nov/20 $$\:{solve}\:\underset{\pi/\mathrm{4}} {\overset{\pi/\mathrm{3}} {\int}}\:\frac{\sqrt{\mathrm{tan}\:{x}}}{\mathrm{sin}\:\mathrm{2}{x}}\:{dx}\:.\: \\ $$$$\:{Solution}\::\: \\ $$$${B}\left({x}\right)=\:\int\:\frac{\sqrt{\mathrm{tan}\:{x}}}{\mathrm{2sin}\:{x}\:\mathrm{cos}\:{x}}\:{dx}\: \\ $$$$\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int\:\frac{{dx}}{\:\sqrt{\mathrm{sin}\:{x}}\:\mathrm{cos}\:{x}\:\sqrt{\mathrm{cos}\:{x}}}…
Question Number 122823 by bemath last updated on 19/Nov/20 Answered by liberty last updated on 19/Nov/20 $$\:{L}\left({x}\right)\:=\:\int\:\frac{\mathrm{1}}{\:\sqrt{{x}\left(\mathrm{1}+\sqrt{{x}}\right)}}\:{dx}\:=\:\int\:\frac{\mathrm{1}}{\:\sqrt{{x}}\:\sqrt{\mathrm{1}+\sqrt{{x}}}\:}\:{dx} \\ $$$${let}\:{u}\:=\:\mathrm{1}+\sqrt{{x}}\:\Rightarrow{du}\:=\:\frac{{dx}}{\mathrm{2}\sqrt{{x}}} \\ $$$${L}\left({x}\right)\:=\:\mathrm{2}\int\:\frac{\mathrm{1}}{\:\sqrt{{u}}}\:{du}\:=\:\mathrm{4}\sqrt{{u}}\:+\:{c}\:\: \\ $$$${L}\left({x}\right)=\:\mathrm{4}\sqrt{\mathrm{1}+\sqrt{{x}}}\:+\:{c}\: \\ $$…
Question Number 122822 by bemath last updated on 19/Nov/20 Answered by liberty last updated on 20/Nov/20 $${I}=\int\:{x}\:\mathrm{tan}\:{x}\:\mathrm{sec}\:^{\mathrm{2}} {x}\:{dx}\:=\:\int\:{x}\:\mathrm{tan}\:{x}\:{d}\left(\mathrm{tan}\:{x}\right) \\ $$$${by}\:{D}.{I}\:{method}\:\rightarrow\begin{cases}{{u}={x}\rightarrow{du}={dx}}\\{{v}=\int\mathrm{tan}\:{x}\:{d}\left(\mathrm{tan}\:{x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{tan}\:^{\mathrm{2}} {x}}\end{cases} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{2}}{x}.\mathrm{tan}\:^{\mathrm{2}} {x}−\frac{\mathrm{1}}{\mathrm{2}}\int\mathrm{tan}\:^{\mathrm{2}} {x}\:{dx}\:…
Question Number 122807 by TITA last updated on 19/Nov/20 Commented by TITA last updated on 19/Nov/20 $${please}\:{help} \\ $$ Answered by ebi last updated on…
Question Number 57241 by Aditya789 last updated on 01/Apr/19 $$\int\frac{×\sqrt{\mathrm{x}+\mathrm{1}}}{\mathrm{x}+\mathrm{2}}\mathrm{dx} \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 01/Apr/19 $${x}+\mathrm{1}={t}^{\mathrm{2}} \rightarrow{dx}=\mathrm{2}{tdt} \\ $$$$\int\frac{\left({t}^{\mathrm{2}} −\mathrm{1}\right){t}×\mathrm{2}{tdt}}{{t}^{\mathrm{2}} +\mathrm{1}}…