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Question Number 36217 Answers: 1 Comments: 0

Suppose a_1 ,...,a_n ,are non−negative reals such that S= a_1 +...+a_n < proof that 1 + S≤ (1 + a_1 )._(...) .(1+ a_n ) ≤ (1/(1−s))

$$\mathrm{Suppose}\:{a}_{\mathrm{1}} ,...,{a}_{{n}} ,\mathrm{are}\:\mathrm{non}−\mathrm{negative} \\ $$$$\mathrm{reals}\:\mathrm{such}\:\mathrm{that}\:{S}=\:{a}_{\mathrm{1}} +...+{a}_{{n}} < \\ $$$${proof}\:{that}\: \\ $$$$\mathrm{1}\:+\:\mathrm{S}\leqslant\:\left(\mathrm{1}\:+\:{a}_{\mathrm{1}} \right)._{...} .\left(\mathrm{1}+\:{a}_{{n}} \right)\:\leqslant\:\frac{\mathrm{1}}{\mathrm{1}−\mathrm{s}} \\ $$

Question Number 36207 Answers: 0 Comments: 4

x^4 +10x^3 +6x−1=0 for those who want an exact solution... (x−a−bi)(x−a+bi)(x−c−d)(x−c+d)=0 x^4 −2(a+c)x^3 +(a^2 +4ac+b^2 +c^2 −d^2 )x^2 − −2(a(c^2 −d^2 )+c(a^2 +b^2 ))x+(a^2 +b^2 )(c^2 −d^2 )=0 1. −2(a+c)=10 2. a^2 +4ac+b^2 +c^2 −d^2 =0 3. −2(a(c^2 −d^2 )+c(a^2 +b^2 ))=6 4. (a^2 +b^2 )(c^2 −d^2 )=−1 1. a=−c−5e 2. b=(√(−a^2 +4ac−c^2 +d^2 ))=(√(2c^2 +10c+d^2 −25)) 3. d=(√(ac+c^2 +((b^2 c+3)/a)))=(√(−((2c^3 +15c^2 +3)/(2c+5)))) 2. b=(√((2c^3 +15c^2 −128)/(2c+5))) 4. 4c^4 +40c^3 +75c^2 −125c+((689)/4)−((17161)/(4(2c+5)^2 ))=0 64(c^6 +15c^5 +75c^4 +125c^3 +4c^2 +20c+1)=0 c^6 +15c^5 +75c^4 +125c^3 +4c^2 +20c+1=0 c=u−((15)/6) u^6 −((75)/4)u^4 +((1939)/(16))u^2 −((17161)/(64))=0 u=(√v) v^3 −((75)/4)v^2 +((1939)/(16))v−((17161)/(64))=0 v=w+((75)/(12)) w^3 +4w+1=0 w=((−(q/2)+(√((p^3 /(27))+(q^2 /4)))))^(1/3) +((−(q/2)−(√((p^3 /(27))+(q^2 /4)))))^(1/3) = =((−(1/2)+((√(849))/(18))))^(1/3) +((−(1/2)−((√(849))/(18))))^(1/3) and now we have to go all the way back... not very friendly... as I mentioned before, it′s almost (or maybe absolutely) impossible to find a nicer exact form of w, so let me use the approximation w≈−.246266 v=((25)/4)+w≈6.00373 u=((√(25+4w))/2)≈2.45025 c=−(5/2)+((√(25+4w))/2)≈−.0497482 d=(√(((25)/2)−w−((131)/(2(√(25+4w))))))≈.787215i a=−(5/2)−((√(25+4w))/2)≈−4.95025 b=(√(−((25)/2)+w+((131)/(2(√(25+4w))))))≈5.11001i x_1 =a+bi≈−10.0603 x_2 =a−bi≈.159762 x_3 =c+d≈−.0497482+.787215i x_4 =c−d≈−.0497482−.787215i

$${x}^{\mathrm{4}} +\mathrm{10}{x}^{\mathrm{3}} +\mathrm{6}{x}−\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{for}\:\mathrm{those}\:\mathrm{who}\:\mathrm{want}\:\mathrm{an}\:\mathrm{exact}\:\mathrm{solution}... \\ $$$$ \\ $$$$\left({x}−{a}−{b}\mathrm{i}\right)\left({x}−{a}+{b}\mathrm{i}\right)\left({x}−{c}−{d}\right)\left({x}−{c}+{d}\right)=\mathrm{0} \\ $$$${x}^{\mathrm{4}} −\mathrm{2}\left({a}+{c}\right){x}^{\mathrm{3}} +\left({a}^{\mathrm{2}} +\mathrm{4}{ac}+{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{d}^{\mathrm{2}} \right){x}^{\mathrm{2}} − \\ $$$$\:\:\:\:\:−\mathrm{2}\left({a}\left({c}^{\mathrm{2}} −{d}^{\mathrm{2}} \right)+{c}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\right){x}+\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\left({c}^{\mathrm{2}} −{d}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$ \\ $$$$\mathrm{1}.\:−\mathrm{2}\left({a}+{c}\right)=\mathrm{10} \\ $$$$\mathrm{2}.\:{a}^{\mathrm{2}} +\mathrm{4}{ac}+{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{d}^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{3}.\:−\mathrm{2}\left({a}\left({c}^{\mathrm{2}} −{d}^{\mathrm{2}} \right)+{c}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\right)=\mathrm{6} \\ $$$$\mathrm{4}.\:\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\left({c}^{\mathrm{2}} −{d}^{\mathrm{2}} \right)=−\mathrm{1} \\ $$$$ \\ $$$$\mathrm{1}.\:{a}=−{c}−\mathrm{5}{e} \\ $$$$\mathrm{2}.\:{b}=\sqrt{−{a}^{\mathrm{2}} +\mathrm{4}{ac}−{c}^{\mathrm{2}} +{d}^{\mathrm{2}} }=\sqrt{\mathrm{2}{c}^{\mathrm{2}} +\mathrm{10}{c}+{d}^{\mathrm{2}} −\mathrm{25}} \\ $$$$\mathrm{3}.\:{d}=\sqrt{{ac}+{c}^{\mathrm{2}} +\frac{{b}^{\mathrm{2}} {c}+\mathrm{3}}{{a}}}=\sqrt{−\frac{\mathrm{2}{c}^{\mathrm{3}} +\mathrm{15}{c}^{\mathrm{2}} +\mathrm{3}}{\mathrm{2}{c}+\mathrm{5}}} \\ $$$$\mathrm{2}.\:{b}=\sqrt{\frac{\mathrm{2}{c}^{\mathrm{3}} +\mathrm{15}{c}^{\mathrm{2}} −\mathrm{128}}{\mathrm{2}{c}+\mathrm{5}}} \\ $$$$\mathrm{4}. \\ $$$$\mathrm{4}{c}^{\mathrm{4}} +\mathrm{40}{c}^{\mathrm{3}} +\mathrm{75}{c}^{\mathrm{2}} −\mathrm{125}{c}+\frac{\mathrm{689}}{\mathrm{4}}−\frac{\mathrm{17161}}{\mathrm{4}\left(\mathrm{2}{c}+\mathrm{5}\right)^{\mathrm{2}} }=\mathrm{0} \\ $$$$\mathrm{64}\left({c}^{\mathrm{6}} +\mathrm{15}{c}^{\mathrm{5}} +\mathrm{75}{c}^{\mathrm{4}} +\mathrm{125}{c}^{\mathrm{3}} +\mathrm{4}{c}^{\mathrm{2}} +\mathrm{20}{c}+\mathrm{1}\right)=\mathrm{0} \\ $$$${c}^{\mathrm{6}} +\mathrm{15}{c}^{\mathrm{5}} +\mathrm{75}{c}^{\mathrm{4}} +\mathrm{125}{c}^{\mathrm{3}} +\mathrm{4}{c}^{\mathrm{2}} +\mathrm{20}{c}+\mathrm{1}=\mathrm{0} \\ $$$${c}={u}−\frac{\mathrm{15}}{\mathrm{6}} \\ $$$${u}^{\mathrm{6}} −\frac{\mathrm{75}}{\mathrm{4}}{u}^{\mathrm{4}} +\frac{\mathrm{1939}}{\mathrm{16}}{u}^{\mathrm{2}} −\frac{\mathrm{17161}}{\mathrm{64}}=\mathrm{0} \\ $$$${u}=\sqrt{{v}} \\ $$$${v}^{\mathrm{3}} −\frac{\mathrm{75}}{\mathrm{4}}{v}^{\mathrm{2}} +\frac{\mathrm{1939}}{\mathrm{16}}{v}−\frac{\mathrm{17161}}{\mathrm{64}}=\mathrm{0} \\ $$$${v}={w}+\frac{\mathrm{75}}{\mathrm{12}} \\ $$$${w}^{\mathrm{3}} +\mathrm{4}{w}+\mathrm{1}=\mathrm{0} \\ $$$${w}=\sqrt[{\mathrm{3}}]{−\frac{{q}}{\mathrm{2}}+\sqrt{\frac{{p}^{\mathrm{3}} }{\mathrm{27}}+\frac{{q}^{\mathrm{2}} }{\mathrm{4}}}}+\sqrt[{\mathrm{3}}]{−\frac{{q}}{\mathrm{2}}−\sqrt{\frac{{p}^{\mathrm{3}} }{\mathrm{27}}+\frac{{q}^{\mathrm{2}} }{\mathrm{4}}}}= \\ $$$$=\sqrt[{\mathrm{3}}]{−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{849}}}{\mathrm{18}}}+\sqrt[{\mathrm{3}}]{−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{849}}}{\mathrm{18}}} \\ $$$$ \\ $$$$\mathrm{and}\:\mathrm{now}\:\mathrm{we}\:\mathrm{have}\:\mathrm{to}\:\mathrm{go}\:\mathrm{all}\:\mathrm{the}\:\mathrm{way}\:\mathrm{back}... \\ $$$$\mathrm{not}\:\mathrm{very}\:\mathrm{friendly}... \\ $$$$\mathrm{as}\:\mathrm{I}\:\mathrm{mentioned}\:\mathrm{before},\:\mathrm{it}'\mathrm{s}\:\mathrm{almost}\:\left(\mathrm{or}\:\mathrm{maybe}\right. \\ $$$$\left.\mathrm{absolutely}\right)\:\mathrm{impossible}\:\mathrm{to}\:\mathrm{find}\:\mathrm{a}\:\mathrm{nicer}\:\mathrm{exact} \\ $$$$\mathrm{form}\:\mathrm{of}\:{w},\:\mathrm{so}\:\mathrm{let}\:\mathrm{me}\:\mathrm{use}\:\mathrm{the}\:\mathrm{approximation} \\ $$$$ \\ $$$${w}\approx−.\mathrm{246266} \\ $$$${v}=\frac{\mathrm{25}}{\mathrm{4}}+{w}\approx\mathrm{6}.\mathrm{00373} \\ $$$${u}=\frac{\sqrt{\mathrm{25}+\mathrm{4}{w}}}{\mathrm{2}}\approx\mathrm{2}.\mathrm{45025} \\ $$$${c}=−\frac{\mathrm{5}}{\mathrm{2}}+\frac{\sqrt{\mathrm{25}+\mathrm{4}{w}}}{\mathrm{2}}\approx−.\mathrm{0497482} \\ $$$${d}=\sqrt{\frac{\mathrm{25}}{\mathrm{2}}−{w}−\frac{\mathrm{131}}{\mathrm{2}\sqrt{\mathrm{25}+\mathrm{4}{w}}}}\approx.\mathrm{787215i} \\ $$$${a}=−\frac{\mathrm{5}}{\mathrm{2}}−\frac{\sqrt{\mathrm{25}+\mathrm{4}{w}}}{\mathrm{2}}\approx−\mathrm{4}.\mathrm{95025} \\ $$$${b}=\sqrt{−\frac{\mathrm{25}}{\mathrm{2}}+{w}+\frac{\mathrm{131}}{\mathrm{2}\sqrt{\mathrm{25}+\mathrm{4}{w}}}}\approx\mathrm{5}.\mathrm{11001i} \\ $$$${x}_{\mathrm{1}} ={a}+{b}\mathrm{i}\approx−\mathrm{10}.\mathrm{0603} \\ $$$${x}_{\mathrm{2}} ={a}−{b}\mathrm{i}\approx.\mathrm{159762} \\ $$$${x}_{\mathrm{3}} ={c}+{d}\approx−.\mathrm{0497482}+.\mathrm{787215i} \\ $$$${x}_{\mathrm{4}} ={c}−{d}\approx−.\mathrm{0497482}−.\mathrm{787215i} \\ $$

Question Number 36205 Answers: 0 Comments: 0

calculate ∫_0 ^∞ ((x^2 −1)/((x^2 +1)^2 )) x^(1/3) dx

$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{x}^{\mathrm{2}} −\mathrm{1}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }\:{x}^{\frac{\mathrm{1}}{\mathrm{3}}} \:{dx} \\ $$

Question Number 36204 Answers: 0 Comments: 1

find the value of ∫_0 ^∞ ((x^2 −1)/(x^2 +1)) ((sin(x))/x)dx

$${find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{2}} \:+\mathrm{1}}\:\frac{{sin}\left({x}\right)}{{x}}{dx}\: \\ $$

Question Number 36203 Answers: 0 Comments: 1

let f(t) = ∫_0 ^∞ ((cos(tx))/((2+x^2 )^2 ))dx 1) find a simple form of f(t) 2) calculate ∫_0 ^∞ ((cos(3x))/((2+x^2 )^2 ))dx

$${let}\:{f}\left({t}\right)\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{cos}\left({tx}\right)}{\left(\mathrm{2}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{a}\:{simple}\:{form}\:\:{of}\:{f}\left({t}\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{cos}\left(\mathrm{3}{x}\right)}{\left(\mathrm{2}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx} \\ $$

Question Number 36202 Answers: 0 Comments: 1

calculate ∫_0 ^∞ ((x^2 dx)/((x^2 +1)^3 ))

$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{x}^{\mathrm{2}} {dx}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} } \\ $$

Question Number 36201 Answers: 0 Comments: 1

calculate ∫_0 ^(π/2) (dθ/(1+2sin^2 θ))

$${calculate}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{d}\theta}{\mathrm{1}+\mathrm{2}{sin}^{\mathrm{2}} \theta} \\ $$

Question Number 36200 Answers: 0 Comments: 4

calculate ∫_0 ^(2π) (dθ/((2+cosθ)^2 ))

$${calculate}\:\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\:\frac{{d}\theta}{\left(\mathrm{2}+{cos}\theta\right)^{\mathrm{2}} } \\ $$

Question Number 36198 Answers: 0 Comments: 1

let f(z) = ((z^2 +1)/(z^4 −1)) find (a_(k)) the poles of f and calculate Res(f,a_k )

$${let}\:{f}\left({z}\right)\:=\:\frac{{z}^{\mathrm{2}} \:+\mathrm{1}}{{z}^{\mathrm{4}} −\mathrm{1}} \\ $$$${find}\:\left({a}_{\left.{k}\right)} {the}\:{poles}\:{of}\:{f}\:{and}\:{calculate}\:\right. \\ $$$${Res}\left({f},{a}_{{k}} \right) \\ $$

Question Number 36197 Answers: 0 Comments: 1

find the value of ∫_0 ^(2π) (dx/(cos^2 x +3 sin^2 x))

$${find}\:{the}\:{value}\:{of}\:\:\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\:\:\frac{{dx}}{{cos}^{\mathrm{2}} {x}\:+\mathrm{3}\:{sin}^{\mathrm{2}} {x}} \\ $$

Question Number 36196 Answers: 0 Comments: 0

let ρ>0 and C the circle x^2 +y^2 =ρ^2 calculate ∫_C ydx +xy dy

$${let}\:\rho>\mathrm{0}\:\:{and}\:{C}\:{the}\:{circle}\:{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \:=\rho^{\mathrm{2}} \\ $$$${calculate}\:\int_{{C}} \:{ydx}\:+{xy}\:{dy} \\ $$

Question Number 36195 Answers: 0 Comments: 1

let C ={(x,y)∈R^2 / 0≤x≤1 and y=2x^2 } calculate ∫_C x^2 ydx +(x^2 −y^2 )dy

$${let}\:\:{C}\:=\left\{\left({x},{y}\right)\in{R}^{\mathrm{2}} /\:\mathrm{0}\leqslant{x}\leqslant\mathrm{1}\:{and}\:{y}=\mathrm{2}{x}^{\mathrm{2}} \right\} \\ $$$${calculate}\:\int_{{C}} \:{x}^{\mathrm{2}} {ydx}\:+\left({x}^{\mathrm{2}} \:−{y}^{\mathrm{2}} \right){dy} \\ $$

Question Number 36194 Answers: 0 Comments: 0

let D ={(x,y,z)∈R^2 / 0<z<1 and x^2 +y^2 <z^2 } calculate ∫∫_D xyzdxdydz

$${let}\:{D}\:=\left\{\left({x},{y},{z}\right)\in{R}^{\mathrm{2}} /\:\mathrm{0}<{z}<\mathrm{1}\:{and}\:{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \:<{z}^{\mathrm{2}} \right\} \\ $$$${calculate}\:\int\int_{{D}} {xyzdxdydz} \\ $$

Question Number 36193 Answers: 0 Comments: 1

let D ={(x,y)∈ R^2 / x^2 +y^2 −x<0 and x^2 +y^2 −y >0 and y>0} calculate∫∫_D (x+y)^2 dxdy

$${let}\:{D}\:=\left\{\left({x},{y}\right)\in\:{R}^{\mathrm{2}} \:/\:{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \:−{x}<\mathrm{0}\:{and}\right. \\ $$$$\left.{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \:−{y}\:>\mathrm{0}\:{and}\:{y}>\mathrm{0}\right\} \\ $$$${calculate}\int\int_{{D}} \:\:\left({x}+{y}\right)^{\mathrm{2}} {dxdy} \\ $$

Question Number 36192 Answers: 0 Comments: 1

let D ={(x,y)∈ R^2 /x^2 +y^2 <1} find the value of ∫∫_D ((dxdy)/(x^2 +y^(2 ) + 2))

$${let}\:{D}\:=\left\{\left({x},{y}\right)\in\:{R}^{\mathrm{2}} \:/{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} <\mathrm{1}\right\} \\ $$$${find}\:{the}\:{value}\:{of}\:\int\int_{{D}} \:\frac{{dxdy}}{{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}\:} +\:\mathrm{2}} \\ $$

Question Number 36191 Answers: 0 Comments: 1

let D = {(x,y)∈R^2 /x>0 ,y>0,x+y<1} 1) calculate ∫∫_D ((xy)/(x^2 +y^2 ))dxdy 2) let a>0 ,b>0 calculate ∫∫_D a^x b^y dxdy

$${let}\:{D}\:=\:\left\{\left({x},{y}\right)\in{R}^{\mathrm{2}} \:/{x}>\mathrm{0}\:,{y}>\mathrm{0},{x}+{y}<\mathrm{1}\right\} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:\int\int_{{D}} \:\:\frac{{xy}}{{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} }{dxdy} \\ $$$$\left.\mathrm{2}\right)\:{let}\:{a}>\mathrm{0}\:,{b}>\mathrm{0}\:{calculate}\:\int\int_{{D}} \:{a}^{{x}} {b}^{{y}} {dxdy} \\ $$

Question Number 36190 Answers: 0 Comments: 1

calculate ∫∫_D (x+y)e^(x+y) dxdy with D = {(x,y)∈R^2 / 0<x<2 and 1<y<2 }

$${calculate}\:\:\int\int_{{D}} \left({x}+{y}\right){e}^{{x}+{y}} {dxdy}\:\:{with} \\ $$$${D}\:=\:\left\{\left({x},{y}\right)\in{R}^{\mathrm{2}} \:/\:\mathrm{0}<{x}<\mathrm{2}\:{and}\:\:\mathrm{1}<{y}<\mathrm{2}\:\right\} \\ $$

Question Number 36189 Answers: 0 Comments: 1

let F(x)=∫_0 ^∞ ((e^(−x^2 t) (√t))/(1+t^2 ))dt calculate lim_(x→+∞) F(x) .

$${let}\:{F}\left({x}\right)=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{e}^{−{x}^{\mathrm{2}} {t}} \sqrt{{t}}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$${calculate}\:{lim}_{{x}\rightarrow+\infty} \:{F}\left({x}\right)\:. \\ $$$$ \\ $$

Question Number 36188 Answers: 0 Comments: 1

find the value of ∫_0 ^∞ ((√t)/(1+t^2 ))dt

$${find}\:{the}\:{value}\:{of}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\sqrt{{t}}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$

Question Number 36187 Answers: 0 Comments: 3

let I_n (x)= ∫_0 ^∞ ((t sin(t))/((t^2 +x^2 )^n ))dt 1) find a relation between I_(n+1) and I_n 2) calculate I_2 (x) and I_3 (x) 3) calculate ∫_0 ^∞ ((tsin(t))/((2+t^2 )^2 ))dt

$${let}\:{I}_{{n}} \left({x}\right)=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{t}\:{sin}\left({t}\right)}{\left({t}^{\mathrm{2}} \:+{x}^{\mathrm{2}} \right)^{{n}} }{dt}\: \\ $$$$\left.\mathrm{1}\right)\:{find}\:{a}\:{relation}\:{between}\:{I}_{{n}+\mathrm{1}} \:\:{and}\:{I}_{{n}} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{I}_{\mathrm{2}} \left({x}\right)\:{and}\:{I}_{\mathrm{3}} \left({x}\right)\: \\ $$$$\left.\mathrm{3}\right)\:{calculate}\:\:\int_{\mathrm{0}} ^{\infty} \:\frac{{tsin}\left({t}\right)}{\left(\mathrm{2}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{dt} \\ $$

Question Number 36186 Answers: 0 Comments: 1

find nature of ∫_1 ^(+∞) (√t) sin(t^2 )dt .

$${find}\:{nature}\:{of}\:\int_{\mathrm{1}} ^{+\infty} \sqrt{{t}}\:{sin}\left({t}^{\mathrm{2}} \right){dt}\:. \\ $$

Question Number 36185 Answers: 0 Comments: 2

study the vonvergence of ∫_1 ^(+∞) ((e^(−(1/t)) −cos((1/t)))/t)dt

$${study}\:{the}\:{vonvergence}\:{of}\: \\ $$$$\int_{\mathrm{1}} ^{+\infty} \:\:\frac{{e}^{−\frac{\mathrm{1}}{{t}}} \:−{cos}\left(\frac{\mathrm{1}}{{t}}\right)}{{t}}{dt} \\ $$

Question Number 36184 Answers: 0 Comments: 1

study the convergence of ∫_1 ^(+∞) ((cos(t))/(√t))dt

$${study}\:{the}\:{convergence}\:{of}\:\:\int_{\mathrm{1}} ^{+\infty} \:\frac{{cos}\left({t}\right)}{\sqrt{{t}}}{dt} \\ $$

Question Number 36183 Answers: 0 Comments: 0

calculate ∫_1 ^(+∞) arctan((1/t))dt