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Question Number 61724    Answers: 0   Comments: 6

Σ_(n≥0) n^2 x^n

$$\underset{{n}\geqslant\mathrm{0}} {\sum}{n}^{\mathrm{2}} {x}^{{n}} \\ $$$$ \\ $$

Question Number 61662    Answers: 0   Comments: 1

calculate ∫_(−(π/4)) ^(π/4) ((cosx)/(e^(1/x) +1)) dx

$${calculate}\:\int_{−\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{4}}} \:\:\frac{{cosx}}{{e}^{\frac{\mathrm{1}}{{x}}} \:+\mathrm{1}}\:{dx}\: \\ $$

Question Number 61661    Answers: 0   Comments: 1

1) calculate ∫∫_R^+^2 ((dxdy)/((1+x^2 )(1+y^2 ))) 2) find the value of ∫_0 ^∞ ((ln(x))/(x^2 −1)) dx .

$$\left.\mathrm{1}\right)\:{calculate}\:\int\int_{{R}^{+^{\mathrm{2}} } } \:\:\:\:\:\frac{{dxdy}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{y}^{\mathrm{2}} \right)} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{ln}\left({x}\right)}{{x}^{\mathrm{2}} −\mathrm{1}}\:{dx}\:. \\ $$

Question Number 61660    Answers: 0   Comments: 1

let U_n = ∫_0 ^∞ (dt/((1+t^3 )^n )) dt (n≥1) 1) calculate (U_(n+1) /U_n ) 2) study the serie Σln((U_(n+1) /U_n )) and prove that lim_(n→+∞) U_n =0

$${let}\:{U}_{{n}} =\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{dt}}{\left(\mathrm{1}+{t}^{\mathrm{3}} \right)^{{n}} }\:{dt}\:\:\:\:\left({n}\geqslant\mathrm{1}\right) \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:\frac{{U}_{{n}+\mathrm{1}} }{{U}_{{n}} } \\ $$$$\left.\mathrm{2}\right)\:{study}\:{the}\:{serie}\:\Sigma{ln}\left(\frac{{U}_{{n}+\mathrm{1}} }{{U}_{{n}} }\right)\:\:{and}\:{prove}\:\:{that}\:{lim}_{{n}\rightarrow+\infty} {U}_{{n}} =\mathrm{0} \\ $$

Question Number 61657    Answers: 0   Comments: 0

U_n and V_n are two sequences verify U_n =Σ_(k=0) ^n C_n ^k V_k determine V_n interms of U_k ,0≤k≤n

$${U}_{{n}} \:{and}\:{V}_{{n}} \:\:{are}\:{two}\:{sequences}\:\:{verify}\:\:{U}_{{n}} =\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:{V}_{{k}} \\ $$$${determine}\:{V}_{{n}} \:\:{interms}\:{of}\:\:{U}_{{k}} \:\:\:\:\:\:\:,\mathrm{0}\leqslant{k}\leqslant{n} \\ $$

Question Number 61675    Answers: 1   Comments: 1

Question Number 61674    Answers: 1   Comments: 4

a.∫_( 0) ^( (𝛑/4)) (√(1+tgx)) dx=? b.∫_( 0) ^( 1) (√(1+lnx)) dx=?

$$\mathrm{a}.\underset{\:\:\:\mathrm{0}} {\overset{\:\:\:\:\:\:\:\frac{\boldsymbol{\pi}}{\mathrm{4}}} {\int}}\sqrt{\mathrm{1}+\boldsymbol{\mathrm{tgx}}}\:\boldsymbol{\mathrm{dx}}=? \\ $$$$\mathrm{b}.\underset{\:\:\mathrm{0}} {\overset{\:\:\:\:\:\:\:\:\mathrm{1}} {\int}}\sqrt{\mathrm{1}+\boldsymbol{\mathrm{lnx}}}\:\boldsymbol{\mathrm{dx}}=? \\ $$

Question Number 61654    Answers: 0   Comments: 0

∫_0 ^∞ e^(−e^x ) ln(x) dx = 0.27634

$$\int_{\mathrm{0}} ^{\infty} {e}^{−{e}^{{x}} } {ln}\left({x}\right)\:{dx}\:=\:\mathrm{0}.\mathrm{27634} \\ $$

Question Number 61646    Answers: 0   Comments: 0

let f(x) =e^(−ax) arctan(3x) with a>0 1) calculate f^((n)) (x) and f^((n)) (0) 2) developp f (x) at integr serie . 3) calculate ∫_0 ^∞ f(x)dx .

$${let}\:{f}\left({x}\right)\:={e}^{−{ax}} \:{arctan}\left(\mathrm{3}{x}\right)\:\:\:{with}\:{a}>\mathrm{0} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{f}^{\left({n}\right)} \left({x}\right)\:{and}\:{f}^{\left({n}\right)} \left(\mathrm{0}\right) \\ $$$$\left.\mathrm{2}\right)\:{developp}\:{f}\:\left({x}\right)\:{at}\:{integr}\:{serie}\:. \\ $$$$\left.\mathrm{3}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\infty} \:{f}\left({x}\right){dx}\:. \\ $$

Question Number 61645    Answers: 0   Comments: 1

calculate ∫∫_D ∫(√(x^2 +y^2 +z^2 ))dxdydz with D ={(x,y,z) / 0≤x≤1 ,1≤y≤2 , 2≤z≤3 }

$${calculate}\:\int\int_{{D}} \int\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} }{dxdydz} \\ $$$${with}\:{D}\:=\left\{\left({x},{y},{z}\right)\:/\:\mathrm{0}\leqslant{x}\leqslant\mathrm{1}\:\:,\mathrm{1}\leqslant{y}\leqslant\mathrm{2}\:\:,\:\mathrm{2}\leqslant{z}\leqslant\mathrm{3}\:\right\} \\ $$

Question Number 61652    Answers: 1   Comments: 1

solve inside C z^4 =((1−i)/(1+i(√3)))

$${solve}\:{inside}\:{C}\:\:{z}^{\mathrm{4}} \:=\frac{\mathrm{1}−{i}}{\mathrm{1}+{i}\sqrt{\mathrm{3}}} \\ $$

Question Number 61651    Answers: 0   Comments: 4

let p(x) =(x+i(√3))^n +(x−i(√3))^n with x real 1) simlify p(x) 2) find the roots of P(x) 3)decompose inside C[x] p(x) 4) calculate ∫_0 ^1 p(x)dx

$${let}\:{p}\left({x}\right)\:=\left({x}+{i}\sqrt{\mathrm{3}}\right)^{{n}} +\left({x}−{i}\sqrt{\mathrm{3}}\right)^{{n}} \:\:\:\:{with}\:{x}\:{real} \\ $$$$\left.\mathrm{1}\right)\:{simlify}\:{p}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{find}\:{the}\:{roots}\:{of}\:{P}\left({x}\right) \\ $$$$\left.\mathrm{3}\right){decompose}\:{inside}\:{C}\left[{x}\right]\:\:{p}\left({x}\right) \\ $$$$\left.\mathrm{4}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} {p}\left({x}\right){dx}\: \\ $$

Question Number 61650    Answers: 0   Comments: 4

solve inside N^2 (x+1)(y+2) =2xy

$${solve}\:{inside}\:{N}^{\mathrm{2}} \:\:\:\:\left({x}+\mathrm{1}\right)\left({y}+\mathrm{2}\right)\:=\mathrm{2}{xy} \\ $$

Question Number 61648    Answers: 0   Comments: 1

calculate ∫∫_W (x^2 −2y^2 )(√(x^2 +y^2 +3))dxdy with W ={ (x,y) ∈ R^2 / 1≤x ≤(√3) and x^2 +y^2 −2y ≤ 2 }

$${calculate}\:\int\int_{{W}} \:\left({x}^{\mathrm{2}} −\mathrm{2}{y}^{\mathrm{2}} \right)\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{3}}{dxdy}\:\:\:\:{with} \\ $$$${W}\:=\left\{\:\left({x},{y}\right)\:\in\:{R}^{\mathrm{2}} \:\:/\:\:\:\:\mathrm{1}\leqslant{x}\:\leqslant\sqrt{\mathrm{3}}\:\:{and}\:\:\:{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} −\mathrm{2}{y}\:\leqslant\:\mathrm{2}\:\right\} \\ $$

Question Number 61635    Answers: 1   Comments: 0

2(∫_0 ^( x) y^3 cos xdx)[((yd^2 y)/dx^2 )−((dy/dx))^2 ] = ky^5 sin x ; y(0)=a, y′(0)=0 . solve the differential equation. (Laplace tranforms might be helpful, i think).

$$\mathrm{2}\left(\int_{\mathrm{0}} ^{\:{x}} {y}^{\mathrm{3}} \mathrm{cos}\:{xdx}\right)\left[\frac{{yd}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }−\left(\frac{{dy}}{{dx}}\right)^{\mathrm{2}} \right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\:{ky}^{\mathrm{5}} \mathrm{sin}\:{x}\:\:\:\:\:;\:\: \\ $$$$\:\:{y}\left(\mathrm{0}\right)={a},\:{y}'\left(\mathrm{0}\right)=\mathrm{0}\:. \\ $$$$\:{solve}\:{the}\:{differential}\:{equation}. \\ $$$$\left({Laplace}\:{tranforms}\:{might}\right. \\ $$$$\left.\:\:\:\:{be}\:{helpful},\:{i}\:{think}\right). \\ $$

Question Number 61622    Answers: 0   Comments: 0

If (√(x (√((x+1) (√((x+2) (√((x+3) (√(...)))))))))) = 2019 so (√(x + (√((x+1) + (√((x+2) + (√((x+3) + (√(...)))))))))) = ?

$${If}\:\:\:\sqrt{{x}\:\sqrt{\left({x}+\mathrm{1}\right)\:\sqrt{\left({x}+\mathrm{2}\right)\:\sqrt{\left({x}+\mathrm{3}\right)\:\sqrt{...}}}}}\:\:\:=\:\:\:\mathrm{2019} \\ $$$${so}\:\:\sqrt{{x}\:+\:\sqrt{\left({x}+\mathrm{1}\right)\:+\:\:\sqrt{\left({x}+\mathrm{2}\right)\:+\:\:\sqrt{\left({x}+\mathrm{3}\right)\:+\:\:\sqrt{...}}}}}\:\:\:=\:\:\:? \\ $$$$ \\ $$

Question Number 61667    Answers: 1   Comments: 0

∫(√(tan(x))) dx

$$\int\sqrt{{tan}\left({x}\right)}\:{dx}\: \\ $$

Question Number 61625    Answers: 1   Comments: 0

1+(1/(1+(1/(1+(1/(1+(1/(1+(1/(1+...))))))))))=

$$\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+...}}}}}= \\ $$

Question Number 61614    Answers: 0   Comments: 1

∫_(−1) ^1 (d/dx) (tan^(−1) (1/x))dx =

$$\underset{−\mathrm{1}} {\overset{\mathrm{1}} {\int}}\:\frac{{d}}{{dx}}\:\left(\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{{x}}\right){dx}\:= \\ $$

Question Number 61613    Answers: 1   Comments: 1

S = 1 + (3/(2018)) + (5/(2018^2 )) + (7/(2018^3 )) + ... 4S − S^2 = ?

$${S}\:\:=\:\:\mathrm{1}\:+\:\frac{\mathrm{3}}{\mathrm{2018}}\:+\:\frac{\mathrm{5}}{\mathrm{2018}^{\mathrm{2}} }\:+\:\frac{\mathrm{7}}{\mathrm{2018}^{\mathrm{3}} }\:+\:... \\ $$$$\mathrm{4}{S}\:−\:{S}^{\mathrm{2}} \:\:=\:\:? \\ $$

Question Number 61605    Answers: 0   Comments: 7

solve at Z^2 2x +5y =4

$${solve}\:\:{at}\:{Z}^{\mathrm{2}} \:\:\:\:\mathrm{2}{x}\:+\mathrm{5}{y}\:=\mathrm{4} \\ $$

Question Number 61601    Answers: 1   Comments: 1

calvulate ∫∫_w (x^2 −y^2 )e^(−x−y) dxdy with W={(x,y)∈R^2 /0≤x≤1 and 1≤y≤3}

$${calvulate}\:\int\int_{{w}} \left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right){e}^{−{x}−{y}} {dxdy} \\ $$$${with}\:{W}=\left\{\left({x},{y}\right)\in{R}^{\mathrm{2}} /\mathrm{0}\leqslant{x}\leqslant\mathrm{1}\:{and}\right. \\ $$$$\left.\mathrm{1}\leqslant{y}\leqslant\mathrm{3}\right\} \\ $$

Question Number 61591    Answers: 1   Comments: 8

solve for z∈C (z)^(1/2) =−1 (z)^(1/3) =−1 (z)^(1/4) =−1

$$\mathrm{solve}\:\mathrm{for}\:{z}\in\mathbb{C} \\ $$$$\sqrt[{\mathrm{2}}]{{z}}=−\mathrm{1} \\ $$$$\sqrt[{\mathrm{3}}]{{z}}=−\mathrm{1} \\ $$$$\sqrt[{\mathrm{4}}]{{z}}=−\mathrm{1} \\ $$

Question Number 61569    Answers: 1   Comments: 3

Question Number 61566    Answers: 1   Comments: 0

∫_2 ^4 ((√(ln(9−(6−x)))/((√(ln(9−x))) + (√(ln(3−x))))) dx

$$\int_{\mathrm{2}} ^{\mathrm{4}} \:\frac{\sqrt{{ln}\left(\mathrm{9}−\left(\mathrm{6}−{x}\right)\right.}}{\sqrt{{ln}\left(\mathrm{9}−{x}\right)}\:+\:\sqrt{{ln}\left(\mathrm{3}−{x}\right)}}\:{dx} \\ $$

Question Number 61559    Answers: 2   Comments: 1

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