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Question Number 70873    Answers: 1   Comments: 0

calculate ∫∫_([1,3]^2 ) e^(−x−y) ln(2x+y)dxdy

$${calculate}\:\int\int_{\left[\mathrm{1},\mathrm{3}\right]^{\mathrm{2}} } \:\:\:{e}^{−{x}−{y}} \:{ln}\left(\mathrm{2}{x}+{y}\right){dxdy} \\ $$

Question Number 70872    Answers: 0   Comments: 1

calculate ∫∫_([1,3]^2 ) e^(−x−y) ln(2x+y)dxdy

$${calculate}\:\int\int_{\left[\mathrm{1},\mathrm{3}\right]^{\mathrm{2}} } \:\:\:{e}^{−{x}−{y}} \:{ln}\left(\mathrm{2}{x}+{y}\right){dxdy} \\ $$

Question Number 70871    Answers: 0   Comments: 1

calculate f(x)=∫_(−∞) ^(+∞) ((cos(x(1+t^2 )))/(1+t^2 ))dt with x≥0

$$\:{calculate}\:{f}\left({x}\right)=\int_{−\infty} ^{+\infty} \:\frac{{cos}\left({x}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\:{with}\:{x}\geqslant\mathrm{0} \\ $$

Question Number 70887    Answers: 0   Comments: 2

Question Number 70865    Answers: 0   Comments: 2

Question Number 70870    Answers: 0   Comments: 4

let f(x)=∫_(−∞) ^(+∞) (dt/((t^2 −2t +x^2 )^4 )) with ∣x∣>1 and n integr natural 1)find a explicit form for f(x) 2) determine also g(x)=∫_(−∞) ^(+∞) (dt/((t^2 −2t +x^2 )^5 )) 3)find the values of integrals ∫_(−∞) ^(+∞) (dt/((t^2 −2t +3)^4 )) and ∫_(−∞) ^(+∞) (dt/((t^2 −2t +3)^5 ))

$${let}\:{f}\left({x}\right)=\int_{−\infty} ^{+\infty} \:\:\frac{{dt}}{\left({t}^{\mathrm{2}} −\mathrm{2}{t}\:+{x}^{\mathrm{2}} \right)^{\mathrm{4}} }\:\:{with}\:\:\mid{x}\mid>\mathrm{1} \\ $$$${and}\:{n}\:{integr}\:{natural} \\ $$$$\left.\mathrm{1}\right){find}\:{a}\:{explicit}\:{form}\:{for}\:{f}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{determine}\:{also}\:{g}\left({x}\right)=\int_{−\infty} ^{+\infty} \:\:\frac{{dt}}{\left({t}^{\mathrm{2}} −\mathrm{2}{t}\:+{x}^{\mathrm{2}} \right)^{\mathrm{5}} } \\ $$$$\left.\mathrm{3}\right){find}\:{the}\:{values}\:{of}\:{integrals}\: \\ $$$$\int_{−\infty} ^{+\infty} \:\frac{{dt}}{\left({t}^{\mathrm{2}} −\mathrm{2}{t}\:+\mathrm{3}\right)^{\mathrm{4}} }\:\:{and}\:\int_{−\infty} ^{+\infty} \:\frac{{dt}}{\left({t}^{\mathrm{2}} −\mathrm{2}{t}\:+\mathrm{3}\right)^{\mathrm{5}} } \\ $$

Question Number 70886    Answers: 2   Comments: 11

Question Number 70847    Answers: 2   Comments: 0

Question Number 70842    Answers: 0   Comments: 1

Question Number 70838    Answers: 1   Comments: 0

Question Number 70834    Answers: 1   Comments: 0

Question Number 70831    Answers: 2   Comments: 1

Question Number 70826    Answers: 2   Comments: 1

Question Number 70818    Answers: 1   Comments: 1

what the prove that ∫_a ^b f(x) dx =∫_a ^b f(a+b−x) dx

$$\mathrm{what}\:\mathrm{the}\:\mathrm{prove}\:\mathrm{that} \\ $$$$\int_{\mathrm{a}} ^{\mathrm{b}} \mathrm{f}\left(\mathrm{x}\right)\:\mathrm{dx}\:=\int_{\mathrm{a}} ^{\mathrm{b}} \mathrm{f}\left(\mathrm{a}+\mathrm{b}−\mathrm{x}\right)\:\mathrm{dx} \\ $$

Question Number 70808    Answers: 1   Comments: 0

Question Number 70784    Answers: 1   Comments: 2

Question Number 70783    Answers: 1   Comments: 4

(5/(4∙9)) + 2((9/(16∙25))) + 3(((13)/(36∙49))) + 4(((17)/(64∙81))) + 5(((21)/(100∙121))) + … = ?

$$\frac{\mathrm{5}}{\mathrm{4}\centerdot\mathrm{9}}\:+\:\mathrm{2}\left(\frac{\mathrm{9}}{\mathrm{16}\centerdot\mathrm{25}}\right)\:+\:\mathrm{3}\left(\frac{\mathrm{13}}{\mathrm{36}\centerdot\mathrm{49}}\right)\:+\:\mathrm{4}\left(\frac{\mathrm{17}}{\mathrm{64}\centerdot\mathrm{81}}\right)\:+\:\mathrm{5}\left(\frac{\mathrm{21}}{\mathrm{100}\centerdot\mathrm{121}}\right)\:+\:\ldots\:=\:\:? \\ $$

Question Number 70780    Answers: 0   Comments: 1

(1/(t+(√u)+(√v)))= =(((t−(√u)+(√v))(t+(√u)−(√v))(t−(√u)−(√v)))/((t+(√u)+(√v))(t−(√u)+(√v))(t+(√u)−(√v))(t−(√u)−(√v))))= =((t^3 −((√u)+(√v))t^2 −((√u)−(√v))^2 t+(u−v)((√u)−(√v)))/(t^4 −2(u+v)t^2 +(u−v)^2 )) (1/(t+(u)^(1/3) +(v)^(1/3) ))= determinant (((α=−(1/2)+((√3)/2)i; β=−(1/2)−((√3)/2)i ⇒)),((⇒ α^2 =β; β^2 =α; α+β=−1; αβ=1))) =(((t+α(u)^(1/3) +β(v)^(1/3) )(t+β(u)^(1/3) +α(v)^(1/3) ))/((t+(u)^(1/3) +(v)^(1/3) )(t+α(u)^(1/3) +β(v)^(1/3) )(t+β(u)^(1/3) +α(v)^(1/3) )))= =((t^2 −((u)^(1/3) +(v)^(1/3) )t+(u^2 )^(1/3) −((uv))^(1/3) +(v^2 )^(1/3) )/(t^3 +u+v−3((uv))^(1/3) t))= determinant (((a=t^3 +u+v; b=27uvt^3 )),(((1/(a−(b)^(1/3) ))=(((αa−β(b)^(1/3) )(βa−α(b)^(1/3) ))/((a−(b)^(1/3) )(αa−β(b)^(1/3) )(βa−α(b)^(1/3) )))=)),((=((a^2 +a(b)^(1/3) +(b^2 )^(1/3) )/(a^3 −b))))) =(N/(t^9 +3(u+v)t^6 +3(u^2 −7uv+v^2 )t^3 +(u+v)^3 )) N= =t^8 − −((u)^(1/3) +(v)^(1/3) )t^7 + +((u)^(1/3) +(v)^(1/3) )^2 t^6 + +((u)^(1/3) +(v)^(1/3) )((u)^(1/3) −2(v)^(1/3) )(2(u)^(1/3) −(v)^(1/3) )t^5 − −((u)^(1/3) +(v)^(1/3) )^2 ((u)^(1/3) −2(v)^(1/3) )(2(u)^(1/3) −(v)^(1/3) )t^4 + +((u)^(1/3) +(v)^(1/3) )^3 ((u)^(1/3) −2(v)^(1/3) )(2(u)^(1/3) −(v)^(1/3) )t^3 + +((u^2 )^(1/3) −((uv))^(1/3) +(v^2 )^(1/3) )^3 t^2 − −((u)^(1/3) +(v)^(1/3) )((u^2 )^(1/3) −((uv))^(1/3) +(v^2 )^(1/3) )^3 t+ +((u)^(1/3) +(v)^(1/3) )^2 ((u^2 )^(1/3) −((uv))^(1/3) +(v^2 )^(1/3) )^3 I think there′s no easier way...

$$\frac{\mathrm{1}}{{t}+\sqrt{{u}}+\sqrt{{v}}}= \\ $$$$=\frac{\left({t}−\sqrt{{u}}+\sqrt{{v}}\right)\left({t}+\sqrt{{u}}−\sqrt{{v}}\right)\left({t}−\sqrt{{u}}−\sqrt{{v}}\right)}{\left({t}+\sqrt{{u}}+\sqrt{{v}}\right)\left({t}−\sqrt{{u}}+\sqrt{{v}}\right)\left({t}+\sqrt{{u}}−\sqrt{{v}}\right)\left({t}−\sqrt{{u}}−\sqrt{{v}}\right)}= \\ $$$$=\frac{{t}^{\mathrm{3}} −\left(\sqrt{{u}}+\sqrt{{v}}\right){t}^{\mathrm{2}} −\left(\sqrt{{u}}−\sqrt{{v}}\right)^{\mathrm{2}} {t}+\left({u}−{v}\right)\left(\sqrt{{u}}−\sqrt{{v}}\right)}{{t}^{\mathrm{4}} −\mathrm{2}\left({u}+{v}\right){t}^{\mathrm{2}} +\left({u}−{v}\right)^{\mathrm{2}} } \\ $$$$ \\ $$$$\frac{\mathrm{1}}{{t}+\sqrt[{\mathrm{3}}]{{u}}+\sqrt[{\mathrm{3}}]{{v}}}= \\ $$$$\:\:\:\:\:\begin{vmatrix}{\alpha=−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i};\:\beta=−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\:\Rightarrow}\\{\Rightarrow\:\alpha^{\mathrm{2}} =\beta;\:\beta^{\mathrm{2}} =\alpha;\:\alpha+\beta=−\mathrm{1};\:\alpha\beta=\mathrm{1}}\end{vmatrix} \\ $$$$=\frac{\left({t}+\alpha\sqrt[{\mathrm{3}}]{{u}}+\beta\sqrt[{\mathrm{3}}]{{v}}\right)\left({t}+\beta\sqrt[{\mathrm{3}}]{{u}}+\alpha\sqrt[{\mathrm{3}}]{{v}}\right)}{\left({t}+\sqrt[{\mathrm{3}}]{{u}}+\sqrt[{\mathrm{3}}]{{v}}\right)\left({t}+\alpha\sqrt[{\mathrm{3}}]{{u}}+\beta\sqrt[{\mathrm{3}}]{{v}}\right)\left({t}+\beta\sqrt[{\mathrm{3}}]{{u}}+\alpha\sqrt[{\mathrm{3}}]{{v}}\right)}= \\ $$$$=\frac{{t}^{\mathrm{2}} −\left(\sqrt[{\mathrm{3}}]{{u}}+\sqrt[{\mathrm{3}}]{{v}}\right){t}+\sqrt[{\mathrm{3}}]{{u}^{\mathrm{2}} }−\sqrt[{\mathrm{3}}]{{uv}}+\sqrt[{\mathrm{3}}]{{v}^{\mathrm{2}} }}{{t}^{\mathrm{3}} +{u}+{v}−\mathrm{3}\sqrt[{\mathrm{3}}]{{uv}}\:{t}}= \\ $$$$\:\:\:\:\:\begin{vmatrix}{{a}={t}^{\mathrm{3}} +{u}+{v};\:{b}=\mathrm{27}{uvt}^{\mathrm{3}} }\\{\frac{\mathrm{1}}{{a}−\sqrt[{\mathrm{3}}]{{b}}}=\frac{\left(\alpha{a}−\beta\sqrt[{\mathrm{3}}]{{b}}\right)\left(\beta{a}−\alpha\sqrt[{\mathrm{3}}]{{b}}\right)}{\left({a}−\sqrt[{\mathrm{3}}]{{b}}\right)\left(\alpha{a}−\beta\sqrt[{\mathrm{3}}]{{b}}\right)\left(\beta{a}−\alpha\sqrt[{\mathrm{3}}]{{b}}\right)}=}\\{=\frac{{a}^{\mathrm{2}} +{a}\sqrt[{\mathrm{3}}]{{b}}+\sqrt[{\mathrm{3}}]{{b}^{\mathrm{2}} }}{{a}^{\mathrm{3}} −{b}}}\end{vmatrix} \\ $$$$=\frac{{N}}{{t}^{\mathrm{9}} +\mathrm{3}\left({u}+{v}\right){t}^{\mathrm{6}} +\mathrm{3}\left({u}^{\mathrm{2}} −\mathrm{7}{uv}+{v}^{\mathrm{2}} \right){t}^{\mathrm{3}} +\left({u}+{v}\right)^{\mathrm{3}} } \\ $$$${N}= \\ $$$$={t}^{\mathrm{8}} − \\ $$$$−\left(\sqrt[{\mathrm{3}}]{{u}}+\sqrt[{\mathrm{3}}]{{v}}\right){t}^{\mathrm{7}} + \\ $$$$+\left(\sqrt[{\mathrm{3}}]{{u}}+\sqrt[{\mathrm{3}}]{{v}}\right)^{\mathrm{2}} {t}^{\mathrm{6}} + \\ $$$$+\left(\sqrt[{\mathrm{3}}]{{u}}+\sqrt[{\mathrm{3}}]{{v}}\right)\left(\sqrt[{\mathrm{3}}]{{u}}−\mathrm{2}\sqrt[{\mathrm{3}}]{{v}}\right)\left(\mathrm{2}\sqrt[{\mathrm{3}}]{{u}}−\sqrt[{\mathrm{3}}]{{v}}\right){t}^{\mathrm{5}} − \\ $$$$−\left(\sqrt[{\mathrm{3}}]{{u}}+\sqrt[{\mathrm{3}}]{{v}}\right)^{\mathrm{2}} \left(\sqrt[{\mathrm{3}}]{{u}}−\mathrm{2}\sqrt[{\mathrm{3}}]{{v}}\right)\left(\mathrm{2}\sqrt[{\mathrm{3}}]{{u}}−\sqrt[{\mathrm{3}}]{{v}}\right){t}^{\mathrm{4}} + \\ $$$$+\left(\sqrt[{\mathrm{3}}]{{u}}+\sqrt[{\mathrm{3}}]{{v}}\right)^{\mathrm{3}} \left(\sqrt[{\mathrm{3}}]{{u}}−\mathrm{2}\sqrt[{\mathrm{3}}]{{v}}\right)\left(\mathrm{2}\sqrt[{\mathrm{3}}]{{u}}−\sqrt[{\mathrm{3}}]{{v}}\right){t}^{\mathrm{3}} + \\ $$$$+\left(\sqrt[{\mathrm{3}}]{{u}^{\mathrm{2}} }−\sqrt[{\mathrm{3}}]{{uv}}+\sqrt[{\mathrm{3}}]{{v}^{\mathrm{2}} }\right)^{\mathrm{3}} {t}^{\mathrm{2}} − \\ $$$$−\left(\sqrt[{\mathrm{3}}]{{u}}+\sqrt[{\mathrm{3}}]{{v}}\right)\left(\sqrt[{\mathrm{3}}]{{u}^{\mathrm{2}} }−\sqrt[{\mathrm{3}}]{{uv}}+\sqrt[{\mathrm{3}}]{{v}^{\mathrm{2}} }\right)^{\mathrm{3}} {t}+ \\ $$$$+\left(\sqrt[{\mathrm{3}}]{{u}}+\sqrt[{\mathrm{3}}]{{v}}\right)^{\mathrm{2}} \left(\sqrt[{\mathrm{3}}]{{u}^{\mathrm{2}} }−\sqrt[{\mathrm{3}}]{{uv}}+\sqrt[{\mathrm{3}}]{{v}^{\mathrm{2}} }\right)^{\mathrm{3}} \\ $$$$ \\ $$$$\mathrm{I}\:\mathrm{think}\:\mathrm{there}'\mathrm{s}\:\mathrm{no}\:\mathrm{easier}\:\mathrm{way}... \\ $$

Question Number 70768    Answers: 1   Comments: 1

Question Number 70757    Answers: 0   Comments: 3

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Question Number 70756    Answers: 0   Comments: 1

I cannot understand all those who post questions like ∫x^(Γ(x^2 )) cos ((log_(ϖ+x) x^(2πi) ))^(1/x) dx=? and a few minutes later (5/3)×((2+1)/(9−4))=? I mean, are you serious?

$$\mathrm{I}\:\mathrm{cannot}\:\mathrm{understand}\:\mathrm{all}\:\mathrm{those}\:\mathrm{who}\:\mathrm{post} \\ $$$$\mathrm{questions}\:\mathrm{like}\:\int{x}^{\Gamma\left({x}^{\mathrm{2}} \right)} \mathrm{cos}\:\sqrt[{{x}}]{\mathrm{log}_{\varpi+{x}} \:{x}^{\mathrm{2}\pi\mathrm{i}} }{dx}=? \\ $$$$\mathrm{and}\:\mathrm{a}\:\mathrm{few}\:\mathrm{minutes}\:\mathrm{later}\:\frac{\mathrm{5}}{\mathrm{3}}×\frac{\mathrm{2}+\mathrm{1}}{\mathrm{9}−\mathrm{4}}=? \\ $$$$\mathrm{I}\:\mathrm{mean},\:\mathrm{are}\:\mathrm{you}\:\mathrm{serious}? \\ $$

Question Number 70782    Answers: 0   Comments: 3

how can i get my password ? i can′t remeber

$$\mathrm{how}\:\mathrm{can}\:\mathrm{i}\:\mathrm{get}\:\mathrm{my}\:\mathrm{password}\:?\:\mathrm{i}\:\:\mathrm{can}'\mathrm{t}\:\mathrm{remeber} \\ $$

Question Number 70746    Answers: 0   Comments: 4

Question Number 70742    Answers: 0   Comments: 1

Prove that f(x) = x − a cos (x) − b has at least one real root for ∀a,b ∈ R

$$\mathrm{Prove}\:\mathrm{that}\:{f}\left({x}\right)\:=\:{x}\:−\:{a}\:\mathrm{cos}\:\left({x}\right)\:−\:{b} \\ $$$$\mathrm{has}\:\mathrm{at}\:\mathrm{least}\:\mathrm{one}\:\mathrm{real}\:\mathrm{root}\:\mathrm{for}\:\forall{a},{b}\:\in\:\mathbb{R} \\ $$

Question Number 70738    Answers: 1   Comments: 2

Question Number 70741    Answers: 0   Comments: 5

f(x)= { ((x^2 x≤1)),(),((∣x−2∣ x>1)) :} find the critical points

$${f}\left({x}\right)=\begin{cases}{{x}^{\mathrm{2}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}\leqslant\mathrm{1}}\\{}\\{\mid{x}−\mathrm{2}\mid\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}>\mathrm{1}}\end{cases} \\ $$$$ \\ $$$${find}\:{the}\:{critical}\:{points} \\ $$

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