Question Number 182367 by universe last updated on 08/Dec/22 $$ \\ $$$$\:\:\:\:\mathrm{find}\:\mathrm{volume}\:\mathrm{of}\:\:\mathrm{region}\:\mathrm{bounded}\:\mathrm{above}\: \\ $$$$\:\:\:\mathrm{by}\:\mathrm{z}\:=\:\mathrm{1}+\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} \:}\:\:\mathrm{and}\:\mathrm{below} \\ $$$$\:\:\:\:\mathrm{by}\:\:\:\mathrm{z}\:=\:\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \:}\: \\ $$ Answered by Tokugami…
Question Number 116815 by bemath last updated on 07/Oct/20 $$\int\:\frac{\mathrm{dx}}{\left(\mathrm{x}−\mathrm{2}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{4}\right)}\:=? \\ $$ Answered by john santu last updated on 07/Oct/20 $$\Rightarrow\:\frac{\mathrm{1}}{\left({x}−\mathrm{2}\right)\left({x}^{\mathrm{2}} +\mathrm{4}\right)}\:=\:\frac{{A}}{{x}−\mathrm{2}}\:+\:\frac{{Bx}+{C}}{{x}^{\mathrm{2}} +\mathrm{4}} \\…
Question Number 116813 by bemath last updated on 07/Oct/20 $$\:\:\:\:\:\:\underset{\mathrm{1}} {\overset{\sqrt{\mathrm{3}}} {\int}}\:\frac{\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }}{\mathrm{x}^{\mathrm{2}} }\:\mathrm{dx}\:? \\ $$ Answered by bobhans last updated on 07/Oct/20 $$\:\mathrm{letting}\:\mathrm{x}\:=\:\mathrm{tan}\:\theta\:\rightarrow\begin{cases}{\theta=\frac{\pi}{\mathrm{3}}}\\{\theta=\frac{\pi}{\mathrm{4}}}\end{cases} \\…
Question Number 116806 by Backer last updated on 07/Oct/20 $$\mathrm{Hi} \\ $$$$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\:\int_{-\infty} ^{+\infty} -\mathrm{e}^{-\mathrm{x}^{\mathrm{2}} } \mathrm{dx}=\sqrt{\pi} \\ $$$$\mathrm{Thanks}\:\mathrm{beforehand} \\ $$$$ \\ $$ Answered…
Question Number 116803 by Ar Brandon last updated on 08/Oct/20 $$\mathrm{Solve}\:\mathrm{for}\:\mathrm{X}\left(\mathrm{x},\mathrm{y},\mathrm{z}\right),\:\mathrm{Y}\left(\mathrm{x},\mathrm{y},\mathrm{z}\right),\:\mathrm{Z}\left(\mathrm{x},\mathrm{y},\mathrm{z}\right) \\ $$$$\begin{cases}{\frac{\partial\mathrm{Z}}{\partial\mathrm{y}}−\frac{\partial\mathrm{Y}}{\partial\mathrm{z}}=\mathrm{1}−\mathrm{x}^{\mathrm{2}} }\\{\frac{\partial\mathrm{Z}}{\partial\mathrm{x}}−\frac{\partial\mathrm{X}}{\partial\mathrm{z}}=−\frac{\mathrm{y}^{\mathrm{2}} }{\mathrm{2}}}\\{\frac{\partial\mathrm{Y}}{\partial\mathrm{x}}−\frac{\partial\mathrm{X}}{\partial\mathrm{y}}=\mathrm{z}\left(\mathrm{2x}−\mathrm{y}\right)}\end{cases}\:\mathrm{where}\:\begin{cases}{\mathrm{X}\left(\mathrm{x},\mathrm{y},\mathrm{0}\right)=\mathrm{0}}\\{\mathrm{Y}\left(\mathrm{x},\mathrm{y},\mathrm{0}\right)=\mathrm{0}}\\{\mathrm{Z}\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)=\mathrm{0}}\end{cases} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 116796 by mnjuly1970 last updated on 08/Oct/20 $$\:\:\:\:\:\:\:\:…\:\:\:\:\:\:{calculus}\:\:… \\ $$$$ \\ $$$$\:\:\:\:\:{prove}\:\:{that}\::: \\ $$$$ \\ $$$$\:\:\:\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\left(\mathrm{1}−{x}^{{p}} \right)\left(\mathrm{1}−{x}^{{q}} \right){x}^{{r}−\mathrm{1}} }{{log}\left({x}\right)}{dx}\overset{???} {=}{log}\left(\:\frac{\left({p}+{q}+{r}+\mathrm{1}\right){r}}{\left({p}+{r}\right)\left({q}+{r}\right)}\:\right) \\…
Question Number 116763 by Henri Boucatchou last updated on 06/Oct/20 $$ \\ $$$$\:\:\:\:\:\int\frac{{dx}}{\:\sqrt[{\mathrm{3}}]{\mathrm{1}+{x}^{\mathrm{3}} }}=? \\ $$ Commented by Lordose last updated on 08/Oct/20 $$\mathrm{feel}\:\mathrm{like}\:\mathrm{attempting} \\…
Question Number 182295 by universe last updated on 07/Dec/22 Answered by SEKRET last updated on 07/Dec/22 $$\boldsymbol{\mathrm{d}}? \\ $$ Commented by universe last updated on…
Question Number 182291 by Frix last updated on 07/Dec/22 $${I}_{\mathrm{2}} =\underset{\mathrm{0}} {\overset{\infty} {\int}}\left(\frac{\mathrm{tan}^{−\mathrm{1}} \:{x}}{{x}}\right)^{\mathrm{2}} {dx}=?? \\ $$$${I}_{\mathrm{3}} =\underset{\mathrm{0}} {\overset{\infty} {\int}}\left(\frac{\mathrm{tan}^{−\mathrm{1}} \:{x}}{{x}}\right)^{\mathrm{3}} {dx}=??? \\ $$$${I}_{\mathrm{4}} =\underset{\mathrm{0}}…
Question Number 51215 by gunawan last updated on 25/Dec/18 $$\int_{\mathrm{0}} ^{\pi} {e}^{\left(\mathrm{1}+{i}\right){x}} {dx}=… \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 25/Dec/18 $${a}=\mathrm{1}+{i} \\ $$$$\int_{\mathrm{0}}…