Question Number 72789 by Learner-123 last updated on 02/Nov/19 $${Find}\:{the}\:{area}\:{of}\:{the}\:{surface}\:{generated} \\ $$$${by}\:{revolving}\:{the}\:{curve}\:{x}=\frac{{y}^{\mathrm{4}} }{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{8}{y}^{\mathrm{2}} }\: \\ $$$${about}\:{the}\:{x}−{axis}\:.\:\left({given}:\mathrm{1}\leqslant{y}\leqslant\mathrm{2}\right) \\ $$ Commented by MJS last updated on 02/Nov/19…
Question Number 138296 by bobhans last updated on 12/Apr/21 $$\int\:\frac{{dx}}{{x}^{\mathrm{4}} \sqrt{{x}^{\mathrm{2}} −{a}^{\mathrm{2}} }}\:=? \\ $$ Answered by bemath last updated on 12/Apr/21 Answered by mathmax…
Question Number 138283 by mnjuly1970 last updated on 11/Apr/21 $$\:\:\:\:\:\:\:\:\:\:\:……{advanced}\:\:\:………..\:\:{calculus}…… \\ $$$$\:\:{prove}\:{that}:: \\ $$$$\:\:\:\:\:\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right).{arctan}\left({x}\right)}{{x}^{\mathrm{2}} }{dx}= \\ $$$$\:{proof}::: \\ $$$$\:\:\:\boldsymbol{\phi}\underset{\langle{substitution}\rangle} {\overset{{x}={tan}\left(\theta\right)} {=}}\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}}…
Question Number 138277 by mathmax by abdo last updated on 11/Apr/21 $$\mathrm{calculate}\:\int_{\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} \mathrm{e}^{−\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} } \mathrm{sin}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \right)\mathrm{dxdy} \\ $$ Answered by…
Question Number 138276 by mathmax by abdo last updated on 11/Apr/21 $$\left.\mathrm{1}\right)\:\mathrm{calculate}\:\mathrm{A}_{\mathrm{n}} =\int\int_{\left[\mathrm{0},\mathrm{n}\left[^{\mathrm{2}} \right.\right.} \:\:\:\frac{\mathrm{dxdy}}{\left(\mathrm{2x}^{\mathrm{2}} \:+\mathrm{3y}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\left.\mathrm{2}\right)\mathrm{find}\:\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \mathrm{A}_{\mathrm{n}} \\ $$ Terms of…
Question Number 138275 by mathmax by abdo last updated on 11/Apr/21 $$\left.\mathrm{1}\right)\mathrm{calculate}\:\mathrm{U}_{\mathrm{n}} =\int\int_{\left[\frac{\mathrm{1}}{\mathrm{n}},\mathrm{1}\right]^{\mathrm{2}} } \:\:\:\left(\mathrm{2x}+\mathrm{3y}\right)\sqrt{\mathrm{x}^{\mathrm{2}} \:+\mathrm{y}^{\mathrm{2}} }\mathrm{dxdy} \\ $$$$\left.\mathrm{2}\right)\mathrm{find}\:\int\int_{\left.\right]\left.\mathrm{0},\mathrm{1}\right]^{\mathrm{2}} } \left(\mathrm{2x}+\mathrm{3y}\right)\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} }\mathrm{dxdy} \\ $$…
Question Number 138257 by greg_ed last updated on 11/Apr/21 $$\boldsymbol{\mathrm{hi}}\:! \\ $$$$\boldsymbol{\mathrm{calculate}}\::\: \\ $$$$\int\int_{\mathrm{A}} \left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right){dxdy}\:{with}\:\mathrm{A}=\left\{\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\:+\:\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\:\leqslant\:\mathrm{1}\right\} \\ $$ Answered by…
Question Number 7184 by Yozzia last updated on 15/Aug/16 $${Show}\:{that} \\ $$$$\int\frac{\mathrm{2}^{{x}} }{\left(\mathrm{1}+\sqrt{\mathrm{5}}\right)^{{x}} +\left(\mathrm{3}+\sqrt{\mathrm{5}}\right)^{{x}} }{dx}=\frac{\mathrm{1}}{{ln}\left(\mathrm{1}+\sqrt{\mathrm{5}}\right)−{ln}\mathrm{2}}\left({ln}\left[\mathrm{1}+\left(\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}\right)^{{x}} \right]−\left(\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}\right)^{{x}} \right)+{C} \\ $$$$ \\ $$ Terms of Service Privacy…
Question Number 7157 by Tawakalitu. last updated on 13/Aug/16 $${If}\:{a}\:{and}\:{b}\:{are}\:{positive}\:{numbers} \\ $$$${what}\:{is}\:{the}\:{value}\:{of}\: \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\frac{{e}^{{ax}} \:−\:{e}^{{bx}} }{\left(\mathrm{1}\:+\:{e}^{{ax}} \right)\left(\mathrm{1}\:+\:{e}^{{bx}} \right)}\:{dx}\: \\ $$ Answered by Yozzia…
Question Number 138223 by mnjuly1970 last updated on 11/Apr/21 $$\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:…….{nice}\:…\:…\:…\:{calculus}… \\ $$$$\:\:\:{evaluate}\:::: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\Theta}=\underset{{n}=−\infty} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{3}{n}+\mathrm{1}\right)^{\mathrm{3}} }\:=? \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:……………………. \\ $$ Answered by…