Menu Close

Category: None

prove-sec-2-xdx-tanx-

Question Number 140581 by bounhome last updated on 09/May/21 $${prove}\:\:\int{sec}^{\mathrm{2}} {xdx}={tanx} \\ $$ Answered by MJS_new last updated on 09/May/21 $$\mathrm{tan}\:{x}\:=\frac{\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}} \\ $$$$\frac{{d}}{{dx}}\left[\frac{{u}\left({x}\right)}{{v}\left({x}\right)}\right]=\frac{{u}'\left({x}\right){v}\left({x}\right)−{u}\left({x}\right){v}'\left({x}\right)}{\left({v}\left({x}\right)\right)^{\mathrm{2}} } \\…

Find-the-sum-of-n-1-2-n-n-3-

Question Number 140576 by SOMEDAVONG last updated on 09/May/21 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\underset{\mathrm{n}=\mathrm{1}} {\overset{+\propto} {\sum}}\frac{\mathrm{2}^{\mathrm{n}} }{\mathrm{n}^{\mathrm{3}} }=? \\ $$ Answered by Dwaipayan Shikari last updated on 09/May/21 $${Li}_{\mathrm{3}}…

4-11-lt-x-y-lt-3-8-x-y-Z-min-x-y-

Question Number 75033 by naka3546 last updated on 06/Dec/19 $$\frac{\mathrm{4}}{\mathrm{11}}\:<\:\frac{{x}}{{y}}\:<\:\frac{\mathrm{3}}{\mathrm{8}} \\ $$$${x},\:{y}\:\:\in\:\:\mathbb{Z}^{+} \\ $$$${min}\:\left\{{x}+{y}\right\}\:\:=\:\:? \\ $$ Answered by mr W last updated on 06/Dec/19 $$\frac{\mathrm{4}}{\mathrm{11}}<\frac{{x}}{{y}}<\frac{\mathrm{3}}{\mathrm{8}}…

Question-140534

Question Number 140534 by SOMEDAVONG last updated on 09/May/21 Answered by EDWIN88 last updated on 09/May/21 $$\left(\mathrm{i}\right)\:=\:\frac{\mathrm{16x}−\mathrm{24}}{\left(\mathrm{x}−\mathrm{1}\right)\left(\mathrm{x}−\mathrm{3}\right)\left(\mathrm{x}+\mathrm{3}\right)}\:=\:\frac{\mathrm{a}}{\mathrm{x}−\mathrm{1}}+\frac{\mathrm{b}}{\mathrm{x}−\mathrm{3}}+\frac{\mathrm{c}}{\mathrm{x}+\mathrm{3}} \\ $$$$\mathrm{a}\:=\:\left[\frac{\mathrm{16x}−\mathrm{24}}{\left(\mathrm{x}−\mathrm{3}\right)\left(\mathrm{x}+\mathrm{3}\right)}\:\right]_{\mathrm{x}=\mathrm{1}} =\:\frac{−\mathrm{8}}{−\mathrm{8}}\:=\mathrm{1} \\ $$$$\mathrm{b}=\:\left[\frac{\mathrm{16x}−\mathrm{24}}{\left(\mathrm{x}−\mathrm{1}\right)\left(\mathrm{x}+\mathrm{3}\right)}\:\right]_{\mathrm{x}=\mathrm{3}} =\:\frac{\mathrm{48}−\mathrm{24}}{\mathrm{6}.\mathrm{2}}=\mathrm{2} \\ $$$$\mathrm{c}\:=\:\left[\frac{\mathrm{16x}−\mathrm{24}}{\left(\mathrm{x}−\mathrm{1}\right)\left(\mathrm{x}−\mathrm{3}\right)}\:\right]_{\mathrm{x}=−\mathrm{3}}…

Question-140529

Question Number 140529 by SOMEDAVONG last updated on 09/May/21 Answered by Rasheed.Sindhi last updated on 09/May/21 $$\frac{\mathrm{1}}{{x}^{\mathrm{2}} \left({x}+\mathrm{2}\right)}−\frac{\mathrm{5}}{\left({x}−\mathrm{2}\right)\left({x}+\mathrm{2}\right)}−\frac{\mathrm{4}}{{x}−\mathrm{2}} \\ $$$$=\frac{\mathrm{1}\left({x}−\mathrm{2}\right)−\mathrm{5}\left({x}^{\mathrm{2}} \right)−\mathrm{4}\left(\:{x}^{\mathrm{2}} \left({x}+\mathrm{2}\right)\:\right)}{{x}^{\mathrm{2}} \left({x}−\mathrm{2}\right)\left({x}+\mathrm{2}\right)} \\ $$$$=\frac{{x}−\mathrm{2}−\mathrm{5}{x}^{\mathrm{2}}…

If-the-zeta-function-of-2-is-2-n-1-1-n-2-2-2-6-the-sum-of-infinite-rational-numbers-why-converges-for-2-6-an-irra

Question Number 9458 by geovane10math last updated on 09/Dec/16 $$\mathrm{If}\:\mathrm{the}\:\mathrm{zeta}\:\mathrm{function}\:\mathrm{of}\:\mathrm{2}\:\mathrm{is} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\zeta}\left(\mathrm{2}\right)\:=\:\underset{\boldsymbol{{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\boldsymbol{{n}}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\zeta}\left(\mathrm{2}\right)\:=\:\frac{\boldsymbol{\pi}^{\mathrm{2}} }{\mathrm{6}} \\ $$$$\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{infinite}\:\boldsymbol{\mathrm{rational}}\:\mathrm{numbers}, \\ $$$$\mathrm{why}\:\mathrm{converges}\:\mathrm{for}\:\frac{\boldsymbol{\pi}^{\mathrm{2}} }{\mathrm{6}},\:\mathrm{an}\:\boldsymbol{\mathrm{irrational}} \\ $$$$\mathrm{number}?…

n-N-and-k-N-Given-0-k-n-1-Show-that-1-n-ln-1-k-n-1-k-n-1-k-1-n-lnx-dn-1-n-ln-1-k-1-n-

Question Number 140513 by mathocean1 last updated on 08/May/21 $$\mathrm{n}\:\in\:\mathbb{N}^{\ast} \:\mathrm{and}\:\mathrm{k}\:\in\:\mathbb{N}^{\ast} . \\ $$$$\mathrm{Given}\:\mathrm{0}\leqslant\mathrm{k}\leqslant\mathrm{n}−\mathrm{1}. \\ $$$$\mathrm{Show}\:\mathrm{that}\: \\ $$$$\frac{\mathrm{1}}{\mathrm{n}}\mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{k}}{\mathrm{n}}\right)\leqslant\underset{\mathrm{1}+\frac{\mathrm{k}}{\mathrm{n}}} {\overset{\mathrm{1}+\frac{\mathrm{k}+\mathrm{1}}{\mathrm{n}}} {\int}}\mathrm{lnx}\:\mathrm{dn}\leqslant\frac{\mathrm{1}}{\mathrm{n}}\mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{k}+\mathrm{1}}{\mathrm{n}}\right) \\ $$ Terms of Service…