Question Number 141222 by mathmax by abdo last updated on 16/May/21 $$\mathrm{calculate}\:\int_{\mathrm{0}} ^{\mathrm{4}\pi} \:\:\frac{\mathrm{d}\theta}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{2x}\:\mathrm{cos}\theta\:+\mathrm{1}\right)^{\mathrm{2}} } \\ $$ Answered by Ar Brandon last updated on…
Question Number 10147 by ridwan balatif last updated on 27/Jan/17 $$\int\sqrt{\frac{\mathrm{p}}{\mathrm{p}−\mathrm{1}}}\mathrm{dp}=…? \\ $$ Commented by prakash jain last updated on 27/Jan/17 $${p}={x}^{\mathrm{2}} \\ $$$${dp}=\mathrm{2}{xdx} \\…
Question Number 141189 by Eric002 last updated on 16/May/21 $${find}\:{the}\:{area}\:{of}\:{the}\:{shaded}\:{region} \\ $$$${shown}\:{below}\:{which}\:{is}\:{boinded}\:{by}\:{to}\:{functions}\: \\ $$$${f}\left({x}\right)={x}^{\mathrm{2}} \:\:{g}\left({x}\right)=\mathrm{2}−{x}\:{and}\:{the}\:{x}-{axis} \\ $$$$ \\ $$ Commented by Eric002 last updated on…
Question Number 141191 by Eric002 last updated on 17/May/21 $${find}\:{the}\:{volume}\:{of}\:{the}\:{solid}\:{generated} \\ $$$${when}\:{the}\:{region}\:{bounded}\:{by}\:{the}\:{y}={x} \\ $$$${y}={x}+\mathrm{2},\:{x}=\mathrm{2}\:{and}\:{x}=\mathrm{4}\:{revolved}\:{about}\:{the}\:{x}-{axis} \\ $$ Answered by ajfour last updated on 17/May/21 $$\:{V}=\pi\int_{\mathrm{2}} ^{\:\:\mathrm{4}}…
Question Number 10101 by Tawakalitu ayo mi last updated on 23/Jan/17 $$\int\:\frac{\mathrm{z}\:+\:\mathrm{2}}{\mathrm{10z}\:+\:\mathrm{28}}\:\mathrm{dz} \\ $$ Commented by prakash jain last updated on 24/Jan/17 $$=\frac{\mathrm{1}}{\mathrm{10}}\int\frac{\mathrm{10}{z}+\mathrm{28}−\mathrm{8}}{\mathrm{10}{z}+\mathrm{28}}{dz} \\ $$$$=\frac{\mathrm{1}}{\mathrm{10}}\int\left[\mathrm{1}−\frac{\mathrm{8}}{\mathrm{10}{z}+\mathrm{28}}\right]{dz}…
Question Number 141142 by iloveisrael last updated on 16/May/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 141133 by mnjuly1970 last updated on 16/May/21 $$\: \\ $$$$\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{I}}:=\int_{\mathrm{0}} ^{\:\infty} \frac{\left({x}^{{n}} −\mathrm{1}\right)\left({x}−\mathrm{1}\right)}{{x}^{{n}+\mathrm{3}} −\mathrm{1}}{dx}=?? \\ $$$$ \\ $$ Answered by mathmax by abdo…
Question Number 141134 by mnjuly1970 last updated on 16/May/21 $$\:\:\:\:\: \\ $$$$\:\:\:{prove}\:{that}:: \\ $$$$\:\:\Phi:=\underset{{m},{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{m}^{\mathrm{2}} +{n}^{\mathrm{2}} }=\:\frac{\pi^{\mathrm{2}} }{\mathrm{24}}+\frac{\pi{ln}\left(\mathrm{2}\right)}{\mathrm{8}} \\ $$$$ \\ $$ Answered…
Question Number 75577 by Tony Lin last updated on 13/Dec/19 $$\int_{−\infty} ^{\infty} {xdx}\:{is}\:{divergent}\:{or}\:{convergent}? \\ $$ Answered by MJS last updated on 13/Dec/19 $$\underset{−\infty} {\overset{+\infty} {\int}}{xdx}=\underset{{r}\rightarrow\infty}…
Question Number 75552 by aliesam last updated on 12/Dec/19 Commented by mathmax by abdo last updated on 12/Dec/19 $${let}\:{I}\:=\int_{−\sqrt{\mathrm{3}}} ^{\mathrm{3}\sqrt{\mathrm{3}}} \:\mid\left({x}−\mathrm{2}\right)^{\mathrm{2}} −\mathrm{4}\mid{dx}\:\:\Rightarrow{I}\:\:=\int_{−\sqrt{\mathrm{3}}} ^{\mathrm{3}\sqrt{\mathrm{3}}} \mid{x}^{\mathrm{2}} −\mathrm{4}{x}\mid\:{dx}…