Question and Answers Forum

AllQuestion and Answers: Page 1403

Question Number 65678 Answers: 1 Comments: 1

calculate ∫ ((3x+1)/((x^2 −4)(x^3 +2x−3)))dx

$${calculate}\:\:\int\:\:\:\:\frac{\mathrm{3}{x}+\mathrm{1}}{\left({x}^{\mathrm{2}} −\mathrm{4}\right)\left({x}^{\mathrm{3}} +\mathrm{2}{x}−\mathrm{3}\right)}{dx} \\ $$

Question Number 65677 Answers: 0 Comments: 0

let S_n =Σ_(k=1) ^n (1/(√(k^2 +k))) 1) find a equivalent of S_n when n→+∞ 2)prove that (S_n ) is convergent.

$${let}\:{S}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{\sqrt{{k}^{\mathrm{2}} +{k}}} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{a}\:{equivalent}\:{of}\:{S}_{{n}} \:{when}\:{n}\rightarrow+\infty \\ $$$$\left.\mathrm{2}\right){prove}\:{that}\:\left({S}_{{n}} \right)\:{is}\:{convergent}. \\ $$

Question Number 65676 Answers: 0 Comments: 1

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

$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−{x}^{\mathrm{2}} −\frac{\mathrm{1}}{{x}^{\mathrm{2}} }} {dx} \\ $$

Question Number 65675 Answers: 0 Comments: 0

1) find ∫ (dx/(√((x^2 +1)(x−2))))

$$\left.\mathrm{1}\right)\:{find}\:\int\:\frac{{dx}}{\sqrt{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}−\mathrm{2}\right)}} \\ $$

Question Number 65674 Answers: 0 Comments: 1

find the value of Σ_(n=2) ^∞ (n/((n+1)^2 (n−1)^3 ))

$${find}\:{the}\:{value}\:{of}\:\sum_{{n}=\mathrm{2}} ^{\infty} \:\frac{{n}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} \left({n}−\mathrm{1}\right)^{\mathrm{3}} } \\ $$

Question Number 65673 Answers: 0 Comments: 1

find Σ_(n=2) ^∞ (((−1)^n )/((n^2 −1)^2 ))

$${find}\:\sum_{{n}=\mathrm{2}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left({n}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} } \\ $$

Question Number 65665 Answers: 0 Comments: 1

calculate ∫_(−2) ^(+∞) (e^(−x) /(√(x+2))) dx

$${calculate}\:\int_{−\mathrm{2}} ^{+\infty} \:\:\frac{{e}^{−{x}} }{\sqrt{{x}+\mathrm{2}}}\:{dx} \\ $$$$ \\ $$

Question Number 65664 Answers: 1 Comments: 0

solve (((√(1−x))−(√(2x+1)))/((√(1−x))+(√(2x+1)))) =((x+1)/3)

$${solve}\:\frac{\sqrt{\mathrm{1}−{x}}−\sqrt{\mathrm{2}{x}+\mathrm{1}}}{\sqrt{\mathrm{1}−{x}}+\sqrt{\mathrm{2}{x}+\mathrm{1}}}\:=\frac{{x}+\mathrm{1}}{\mathrm{3}} \\ $$

Question Number 65724 Answers: 2 Comments: 0

Question Number 65723 Answers: 1 Comments: 0

Question Number 65700 Answers: 0 Comments: 1

Question Number 65651 Answers: 0 Comments: 1

Question Number 65704 Answers: 0 Comments: 1

∫ln^2 xsin(x)dx

$$\int{ln}^{\mathrm{2}} {xsin}\left({x}\right){dx} \\ $$

Question Number 65601 Answers: 0 Comments: 1

1.If y=x^(n−1) log x,then prove that,x^2 (d^2 y/dx^2 )+(3−2n)x(dy/dx)+(n−1)^2 y=0 2.If ((mtan (α−θ))/(cos^2 θ))=((ntan θ)/(cos^2 (α−θ))),then prove that,θ=(1/2)[α−tan^(−1) (((n−m)/(n+m))tan α)]

$$\mathrm{1}.\mathrm{If}\:\boldsymbol{{y}}=\boldsymbol{{x}}^{\boldsymbol{{n}}−\mathrm{1}} \mathrm{log}\:\boldsymbol{{x}},\mathrm{then}\:\mathrm{prove}\:\mathrm{that},\boldsymbol{{x}}^{\mathrm{2}} \frac{\boldsymbol{{d}}^{\mathrm{2}} \boldsymbol{{y}}}{\boldsymbol{{dx}}^{\mathrm{2}} }+\left(\mathrm{3}−\mathrm{2}\boldsymbol{{n}}\right)\boldsymbol{{x}}\frac{\boldsymbol{{dy}}}{\boldsymbol{{dx}}}+\left(\boldsymbol{{n}}−\mathrm{1}\right)^{\mathrm{2}} \boldsymbol{{y}}=\mathrm{0} \\ $$$$\mathrm{2}.\boldsymbol{\mathrm{I}}\mathrm{f}\:\frac{\mathrm{mtan}\:\left(\alpha−\theta\right)}{\mathrm{cos}\:^{\mathrm{2}} \theta}=\frac{{n}\mathrm{tan}\:\theta}{\mathrm{cos}\:^{\mathrm{2}} \left(\alpha−\theta\right)},\mathrm{then}\:\mathrm{prove}\:\mathrm{that},\theta=\frac{\mathrm{1}}{\mathrm{2}}\left[\alpha−\mathrm{tan}^{−\mathrm{1}} \left(\frac{{n}−{m}}{{n}+{m}}\mathrm{tan}\:\alpha\right)\right] \\ $$

Question Number 65593 Answers: 1 Comments: 5

∫_0 ^(1/3) (3x + 1)^5 dx =

$$\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{3}}} \left(\mathrm{3}{x}\:+\:\mathrm{1}\right)^{\mathrm{5}} {dx}\:= \\ $$

Question Number 65592 Answers: 1 Comments: 0

∫_0 ^(1/3) (3x + 1)^5 dx =

$$\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{3}}} \left(\mathrm{3}{x}\:+\:\mathrm{1}\right)^{\mathrm{5}} {dx}\:= \\ $$

Question Number 65587 Answers: 1 Comments: 4

Prove that 1^3 + 2^3 + 3^3 + … + n^3 = (1+2+3+...+n)^2

$${Prove}\:\:{that} \\ $$$$\:\:\:\:\:\mathrm{1}^{\mathrm{3}} \:+\:\mathrm{2}^{\mathrm{3}} \:+\:\mathrm{3}^{\mathrm{3}} \:+\:\ldots\:+\:{n}^{\mathrm{3}} \:\:=\:\:\left(\mathrm{1}+\mathrm{2}+\mathrm{3}+...+{n}\right)^{\mathrm{2}} \\ $$

Question Number 65589 Answers: 0 Comments: 8

Question Number 65581 Answers: 0 Comments: 1

Question Number 65580 Answers: 0 Comments: 0

Question Number 65541 Answers: 0 Comments: 0

we are all students learning mutually from each other we are not at par with great Ramanuj and Hardy...to solve problems related to zeta function

$${we}\:{are}\:{all}\:{students}\:{learning}\:{mutually}\:{from}\:{each}\:{other} \\ $$$${we}\:{are}\:{not}\:{at}\:{par}\:{with}\:{great}\:{Ramanuj}\:{and} \\ $$$${Hardy}...{to}\:{solve}\:{problems}\:{related}\:{to}\:{zeta}\:{function} \\ $$

Question Number 65539 Answers: 0 Comments: 0

Question Number 65569 Answers: 1 Comments: 1

reposting this: x^4 +(1−2a)x^2 −2ax+1=0 solve for a>0∧x>0 a^5 −((23)/8)x^4 −((49)/4)x^3 +((27)/8)x^2 +3x+(9/8)=0 ⇒ 3 real solutions a_1 <a_2 <a_3 we cannot solve exactly (?) so it′s not possible to get all the below solutions exactly, only rough approximations for the following α,β,γ,δ ∈R the possible natures of the zeros: (1) a<a_1 ∨a_2 <a<a_3 ⇒ x=α∨x=β∨x=γ±δi (2) a>a_3 ⇒ x=α∨x=β∨x=γ∨x=δ (3) a_1 <a<a_2 ⇒ x=α±βi∨x=γ±δi (4) a=a_3 ⇒ x=α=β∨x=γ∨x=δ (5) a=a_1 ∨a=a_2 ⇒ x=α=β∨x=γ±δi

$$\mathrm{reposting}\:\mathrm{this}: \\ $$$${x}^{\mathrm{4}} +\left(\mathrm{1}−\mathrm{2}{a}\right){x}^{\mathrm{2}} −\mathrm{2}{ax}+\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{solve}\:\mathrm{for}\:{a}>\mathrm{0}\wedge{x}>\mathrm{0} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:{a}^{\mathrm{5}} −\frac{\mathrm{23}}{\mathrm{8}}{x}^{\mathrm{4}} −\frac{\mathrm{49}}{\mathrm{4}}{x}^{\mathrm{3}} +\frac{\mathrm{27}}{\mathrm{8}}{x}^{\mathrm{2}} +\mathrm{3}{x}+\frac{\mathrm{9}}{\mathrm{8}}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\Rightarrow\:\mathrm{3}\:\mathrm{real}\:\mathrm{solutions}\:{a}_{\mathrm{1}} <{a}_{\mathrm{2}} <{a}_{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{we}\:\mathrm{cannot}\:\mathrm{solve}\:\mathrm{exactly}\:\left(?\right)\:\mathrm{so}\:\mathrm{it}'\mathrm{s}\:\mathrm{not} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{possible}\:\mathrm{to}\:\mathrm{get}\:\mathrm{all}\:\mathrm{the}\:\mathrm{below}\:\mathrm{solutions} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{exactly},\:\mathrm{only}\:\mathrm{rough}\:\mathrm{approximations} \\ $$$$ \\ $$$$\mathrm{for}\:\mathrm{the}\:\mathrm{following}\:\alpha,\beta,\gamma,\delta\:\in\mathbb{R} \\ $$$$\mathrm{the}\:\mathrm{possible}\:\mathrm{natures}\:\mathrm{of}\:\mathrm{the}\:\mathrm{zeros}: \\ $$$$\left(\mathrm{1}\right)\:\:{a}<{a}_{\mathrm{1}} \vee{a}_{\mathrm{2}} <{a}<{a}_{\mathrm{3}} \:\Rightarrow\:{x}=\alpha\vee{x}=\beta\vee{x}=\gamma\pm\delta\mathrm{i} \\ $$$$\left(\mathrm{2}\right)\:\:{a}>{a}_{\mathrm{3}} \:\Rightarrow\:{x}=\alpha\vee{x}=\beta\vee{x}=\gamma\vee{x}=\delta \\ $$$$\left(\mathrm{3}\right)\:\:{a}_{\mathrm{1}} <{a}<{a}_{\mathrm{2}} \:\Rightarrow\:{x}=\alpha\pm\beta\mathrm{i}\vee{x}=\gamma\pm\delta\mathrm{i} \\ $$$$\left(\mathrm{4}\right)\:\:{a}={a}_{\mathrm{3}} \:\Rightarrow\:{x}=\alpha=\beta\vee{x}=\gamma\vee{x}=\delta \\ $$$$\left(\mathrm{5}\right)\:\:{a}={a}_{\mathrm{1}} \vee{a}={a}_{\mathrm{2}} \:\Rightarrow\:{x}=\alpha=\beta\vee{x}=\gamma\pm\delta\mathrm{i} \\ $$

Question Number 65495 Answers: 1 Comments: 0

{ (((a/b)+(b/(a+b))=(√3))),(((b/a)+(a/(a+b))=(√2))) :} [a,b∈R]

$$\begin{cases}{\frac{\boldsymbol{\mathrm{a}}}{\boldsymbol{\mathrm{b}}}+\frac{\boldsymbol{\mathrm{b}}}{\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}}=\sqrt{\mathrm{3}}}\\{\frac{\boldsymbol{\mathrm{b}}}{\boldsymbol{\mathrm{a}}}+\frac{\boldsymbol{\mathrm{a}}}{\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}}=\sqrt{\mathrm{2}}}\end{cases}\:\:\:\left[\boldsymbol{\mathrm{a}},\boldsymbol{\mathrm{b}}\in\boldsymbol{\mathrm{R}}\right] \\ $$

Question Number 65492 Answers: 0 Comments: 0

Question Number 65491 Answers: 1 Comments: 1

Pg 1398 Pg 1399 Pg 1400 Pg 1401 Pg 1402 Pg 1403 Pg 1404 Pg 1405 Pg 1406 Pg 1407

Contact: info@tinkutara.com