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Question Number 218563    Answers: 0   Comments: 0

solve for x ∈ R ∫_0 ^∞ ((sin(xt))/(e^t −1 )) dt = (𝛑/2) coth(𝛑x) − (1/(2x))

$$ \\ $$$$\:\:\:\:\:\:\:\:\:{solve}\:{for}\:\boldsymbol{{x}}\:\in\:\mathbb{R}\: \\ $$$$\:\:\:\:\:\int_{\mathrm{0}} ^{\infty} \:\frac{{sin}\left({xt}\right)}{{e}^{{t}} −\mathrm{1}\:}\:{dt}\:=\:\frac{\boldsymbol{\pi}}{\mathrm{2}}\:{coth}\left(\boldsymbol{\pi}{x}\right)\:−\:\frac{\mathrm{1}}{\mathrm{2}{x}}\:\: \\ $$$$ \\ $$

Question Number 218560    Answers: 0   Comments: 1

Question Number 218557    Answers: 1   Comments: 1

$$\: \\ $$

Question Number 218547    Answers: 1   Comments: 1

Question Number 218543    Answers: 1   Comments: 1

Question Number 218539    Answers: 1   Comments: 0

Solve (∂^2 w/∂t^2 )=c^2 (∂^2 w/∂x^2 ) w(0,t)=f(t) ,lim_(x→∞) w(x,t)=0 (Boundary Condition) w(x,0)=0 , w_t (x,0)=0 (Initial Condition) f(t) { ((sin(t) , t∈[0,2π))),((0 , otherwise)) :}

$${S}\mathrm{olve} \\ $$$$\frac{\partial^{\mathrm{2}} {w}}{\partial{t}^{\mathrm{2}} }={c}^{\mathrm{2}} \frac{\partial^{\mathrm{2}} {w}}{\partial{x}^{\mathrm{2}} } \\ $$$${w}\left(\mathrm{0},{t}\right)={f}\left({t}\right)\:,\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{w}\left({x},{t}\right)=\mathrm{0}\:\left(\mathrm{Boundary}\:\mathrm{Condition}\right) \\ $$$${w}\left({x},\mathrm{0}\right)=\mathrm{0}\:,\:{w}_{{t}} \left({x},\mathrm{0}\right)=\mathrm{0}\:\left(\mathrm{Initial}\:\mathrm{Condition}\right) \\ $$$${f}\left({t}\right)\begin{cases}{\mathrm{sin}\left({t}\right)\:,\:{t}\in\left[\mathrm{0},\mathrm{2}\pi\right)}\\{\mathrm{0}\:,\:\mathrm{otherwise}}\end{cases} \\ $$

Question Number 218532    Answers: 0   Comments: 0

Question Number 218534    Answers: 0   Comments: 0

I Find Fun integral problem! ∫ x^dx −1 =?? note) I already know that integral Solution Try integral problem! (# Product Integral , #Integral)

$$\mathrm{I}\:\mathrm{Find}\:\mathrm{Fun}\:\mathrm{integral}\:\mathrm{problem}! \\ $$$$\int\:\:{x}^{\mathrm{d}{x}} −\mathrm{1}\:=??\: \\ $$$$\left.\mathrm{note}\right)\:\:\mathrm{I}\:\mathrm{already}\:\mathrm{know}\:\mathrm{that}\:\mathrm{integral}\:\mathrm{Solution} \\ $$$$\:\mathrm{Try}\:\mathrm{integral}\:\mathrm{problem}! \\ $$$$\left(#\:\mathrm{Product}\:\mathrm{Integral}\:,\:#\mathrm{Integral}\right) \\ $$

Question Number 218528    Answers: 3   Comments: 0

Question Number 218527    Answers: 1   Comments: 0

Question Number 218526    Answers: 2   Comments: 2

Question Number 218525    Answers: 1   Comments: 0

Question Number 218515    Answers: 0   Comments: 0

Question Number 218504    Answers: 1   Comments: 0

Let a>1 fixed Solve for real numbers the system { ((a^x + 2^x = ay + 2)),((a^y + 2^y = az + 2)),((a^z + 2^z = ax + 2)) :}

$$\mathrm{Let}\:\:\:\mathrm{a}>\mathrm{1}\:\:\:\mathrm{fixed} \\ $$$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}\:\mathrm{the}\:\mathrm{system} \\ $$$$\begin{cases}{\mathrm{a}^{\boldsymbol{\mathrm{x}}} \:\:+\:\:\mathrm{2}^{\boldsymbol{\mathrm{x}}} \:\:=\:\:\mathrm{ay}\:+\:\mathrm{2}}\\{\mathrm{a}^{\boldsymbol{\mathrm{y}}} \:\:+\:\:\mathrm{2}^{\boldsymbol{\mathrm{y}}} \:\:=\:\:\mathrm{az}\:\:+\:\:\mathrm{2}}\\{\mathrm{a}^{\boldsymbol{\mathrm{z}}} \:\:+\:\:\mathrm{2}^{\boldsymbol{\mathrm{z}}} \:\:=\:\:\mathrm{ax}\:\:+\:\:\mathrm{2}}\end{cases}\: \\ $$

Question Number 218494    Answers: 4   Comments: 0

Question Number 218493    Answers: 2   Comments: 0

Question Number 218491    Answers: 5   Comments: 1

Question Number 218485    Answers: 1   Comments: 0

solve the equation 1) X^6 −1=0 2) X^4 +X^2 +1=0

$${solve}\:{the}\:{equation} \\ $$$$\left.\mathrm{1}\right)\:\:\:{X}^{\mathrm{6}} −\mathrm{1}=\mathrm{0} \\ $$$$\left.\mathrm{2}\right)\:\:{X}^{\mathrm{4}} +{X}^{\mathrm{2}} +\mathrm{1}=\mathrm{0} \\ $$

Question Number 218482    Answers: 0   Comments: 1

solve for real x; _( 3(√(((x^5 −5x^4 +10x^3 −10x^2 +5x)/(x^2 −2x+1))+(√(x^4 +4x^2 +4))))=(√(3 )) )

$$ \\ $$$$\:\:\:\:{solve}\:{for}\:{real}\:\boldsymbol{{x}}; \\ $$$$ \\ $$$$\underset{\:\:\:\mathrm{3}\sqrt{\frac{\boldsymbol{{x}}^{\mathrm{5}} −\mathrm{5}\boldsymbol{{x}}^{\mathrm{4}} +\mathrm{10}\boldsymbol{{x}}^{\mathrm{3}} −\mathrm{10}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{5}\boldsymbol{{x}}}{\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{2}\boldsymbol{{x}}+\mathrm{1}}+\sqrt{\boldsymbol{{x}}^{\mathrm{4}} +\mathrm{4}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{4}}}=\sqrt{\mathrm{3}\:}\:\:\:} {\:} \\ $$$$ \\ $$

Question Number 218476    Answers: 1   Comments: 0

Question Number 218475    Answers: 5   Comments: 0

Question Number 218474    Answers: 1   Comments: 0

solve for real x (√(x+(√(x^2 +1)))) = (√2)

$$ \\ $$$$\:\:\:\:{solve}\:{for}\:{real}\:\boldsymbol{{x}} \\ $$$$\:\:\:\:\sqrt{{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}\:=\:\sqrt{\mathrm{2}} \\ $$$$ \\ $$

Question Number 218473    Answers: 2   Comments: 0

Solve the following equation for the real x value; x^4 −4x^3 +6x^2 −4x+2 = (√(2x^4 −8x^3 +12x^2 −8x+5))

$$ \\ $$$$\:{Solve}\:{the}\:{following}\:{equation}\:{for}\:{the}\:{real}\:\boldsymbol{{x}}\:{value}; \\ $$$$\:{x}^{\mathrm{4}} −\mathrm{4}{x}^{\mathrm{3}} +\mathrm{6}{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{2}\:=\:\sqrt{\mathrm{2}{x}^{\mathrm{4}} −\mathrm{8}{x}^{\mathrm{3}} +\mathrm{12}{x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{5}}\:\:\:\: \\ $$$$ \\ $$

Question Number 218462    Answers: 2   Comments: 0

∫(dx/( (√(2x−x^2 +3))))

$$\int\frac{{dx}}{\:\sqrt{\mathrm{2}{x}−{x}^{\mathrm{2}} +\mathrm{3}}} \\ $$

Question Number 218459    Answers: 0   Comments: 2

F^→ (x,y)=−(1/2)ye_1 ^→ +(1/2)xe_2 ^→ ▽^→ ×F^→ (x,y)= determinant ((( e_1 ^→ ),e_2 ^→ ,e_3 ^→ ),(( ∂_x ),( ∂_y ),∂_z ),((−(1/2)y),((1/2)x),0))=0e_1 ^→ −0e_2 ^→ +((1/2)−(−(1/2)))e_3 ^→ ∮_( C) F^→ ∙dl=∮_( C) −(1/2)ydx+(1/2)xdx=∫∫_( R^2 ) e_3 ^→ ∙n^→ dS ∴∮_( C) −(1/2)ydx+(1/2)xdx=∫∫_( R^2 ) dS.....is right...??

$$\overset{\rightarrow} {\boldsymbol{\mathrm{F}}}\left({x},{y}\right)=−\frac{\mathrm{1}}{\mathrm{2}}{y}\overset{\rightarrow} {\boldsymbol{\mathrm{e}}}_{\mathrm{1}} +\frac{\mathrm{1}}{\mathrm{2}}{x}\overset{\rightarrow} {\boldsymbol{\mathrm{e}}}_{\mathrm{2}} \\ $$$$\overset{\rightarrow} {\bigtriangledown}×\overset{\rightarrow} {\boldsymbol{\mathrm{F}}}\left({x},{y}\right)=\begin{vmatrix}{\:\:\:\:\overset{\rightarrow} {\boldsymbol{\mathrm{e}}}_{\mathrm{1}} }&{\overset{\rightarrow} {\boldsymbol{\mathrm{e}}}_{\mathrm{2}} }&{\overset{\rightarrow} {\boldsymbol{\mathrm{e}}}_{\mathrm{3}} }\\{\:\:\:\:\:\partial_{{x}} }&{\:\partial_{{y}} }&{\partial_{{z}} }\\{−\frac{\mathrm{1}}{\mathrm{2}}{y}}&{\frac{\mathrm{1}}{\mathrm{2}}{x}}&{\mathrm{0}}\end{vmatrix}=\mathrm{0}\overset{\rightarrow} {\boldsymbol{\mathrm{e}}}_{\mathrm{1}} −\mathrm{0}\overset{\rightarrow} {\boldsymbol{\mathrm{e}}}_{\mathrm{2}} +\left(\frac{\mathrm{1}}{\mathrm{2}}−\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)\right)\overset{\rightarrow} {\boldsymbol{\mathrm{e}}}_{\mathrm{3}} \\ $$$$\oint_{\:{C}} \:\overset{\rightarrow} {\boldsymbol{\mathrm{F}}}\centerdot\mathrm{d}\boldsymbol{\mathrm{l}}=\oint_{\:{C}} −\frac{\mathrm{1}}{\mathrm{2}}{y}\mathrm{d}{x}+\frac{\mathrm{1}}{\mathrm{2}}{x}\mathrm{d}{x}=\int\int_{\:\mathbb{R}^{\mathrm{2}} } \overset{\rightarrow} {\boldsymbol{\mathrm{e}}}_{\mathrm{3}} \centerdot\overset{\rightarrow} {\boldsymbol{\mathrm{n}}}\:\mathrm{dS} \\ $$$$\therefore\oint_{\:{C}} −\frac{\mathrm{1}}{\mathrm{2}}{y}\mathrm{d}{x}+\frac{\mathrm{1}}{\mathrm{2}}{x}\mathrm{d}{x}=\int\int_{\:\mathbb{R}^{\mathrm{2}} } \:\mathrm{dS}.....\mathrm{is}\:\mathrm{right}...?? \\ $$

Question Number 218456    Answers: 5   Comments: 0

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