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Question Number 118975 Answers: 2 Comments: 0

(3x−5y) dx + (x+y) dy = 0

$$\:\left(\mathrm{3}{x}−\mathrm{5}{y}\right)\:{dx}\:+\:\left({x}+{y}\right)\:{dy}\:=\:\mathrm{0} \\ $$

Question Number 118969 Answers: 3 Comments: 0

Given f(x) = ((sin x+cos x)/(sin x−cos x)) find the value of f ′′(x) + f ′(x) + 1 .

$$\:{Given}\:{f}\left({x}\right)\:=\:\frac{\mathrm{sin}\:{x}+\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}−\mathrm{cos}\:{x}} \\ $$$${find}\:{the}\:{value}\:{of}\: \\ $$$$\:{f}\:''\left({x}\right)\:+\:{f}\:'\left({x}\right)\:+\:\mathrm{1}\:. \\ $$

Question Number 118967 Answers: 0 Comments: 0

lim_(x→∞) ((((x!)/x^x ))^(1/x) )^(((2∙4)/(1∙3))∙((4∙6)/(3∙5))∙∙∙) =? (2/1)∙(2/3)∙(4/3)∙(4/5)∙(6/5)∙(6/7)∙∙∙=(π/2) lim_(x→∞) ((((x!)/x^x ))^(1/x) )^((2/1)∙(4/3)∙(4/3)∙(6/5)∙∙∙) =((1/e))^(π/2) =(√(((1/e))^π ))=(1/( (√e^π )))

Question Number 118962 Answers: 2 Comments: 0

x dy = (y+(√(x^2 +y^2 )) ) dx ; y((√3)) = 1

$$\:{x}\:{dy}\:=\:\left({y}+\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\:\right)\:{dx}\:;\:{y}\left(\sqrt{\mathrm{3}}\right)\:=\:\mathrm{1} \\ $$

Question Number 118960 Answers: 1 Comments: 0

Given f: R→R and g: R→R where f(x)=x^3 +3 and g(x)=2x+1. Find the value of f^(−1) (g^(−1) (23)).

$$\:{Given}\:{f}:\:{R}\rightarrow{R}\:{and}\:{g}:\:{R}\rightarrow{R} \\ $$$${where}\:{f}\left({x}\right)={x}^{\mathrm{3}} +\mathrm{3}\:{and}\:{g}\left({x}\right)=\mathrm{2}{x}+\mathrm{1}.\:{Find} \\ $$$${the}\:{value}\:{of}\:{f}^{−\mathrm{1}} \left({g}^{−\mathrm{1}} \left(\mathrm{23}\right)\right). \\ $$

Question Number 118959 Answers: 2 Comments: 0

∫ (dx/(x^6 −x^3 )) ?

$$\:\int\:\frac{{dx}}{{x}^{\mathrm{6}} −{x}^{\mathrm{3}} }\:? \\ $$

Question Number 118955 Answers: 1 Comments: 0

find ∫_0 ^∞ ((lnx)/(x^2 +x+1))dx

$${find}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{lnx}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}}{dx} \\ $$

Question Number 118954 Answers: 2 Comments: 0

solve (6x^2 + 3y^2 ) dx = 2xy dy

$${solve}\:\left(\mathrm{6}{x}^{\mathrm{2}} \:+\:\mathrm{3}{y}^{\mathrm{2}} \right)\:{dx}\:=\:\mathrm{2}{xy}\:{dy} \\ $$

Question Number 118952 Answers: 2 Comments: 0

Salut Pouvez vous m′aider avec ce devoir? Examiner si les series suivantes sont absolument convergentes ou semi−convergentes Σ_(k=0) ^n (((-1)^(k-1) )/((k+1)^2 )) Σ_(k=0) ^n (((-1)^(k-1) )/k) Etudier la convergence des series suivantes Σ_(k=1) ^n (((2k)!)/2^k ) Σ_(k=1) ^n ((1/2))^k

$$\mathrm{Salut} \\ $$$$\mathrm{Pouvez}\:\mathrm{vous}\:\mathrm{m}'\mathrm{aider}\:\mathrm{avec}\:\mathrm{ce}\:\mathrm{devoir}? \\ $$$$ \\ $$$$\mathrm{Examiner}\:\mathrm{si}\:\mathrm{les}\:\mathrm{series}\:\mathrm{suivantes}\:\mathrm{sont}\: \\ $$$$\mathrm{absolument}\:\mathrm{convergentes}\:\mathrm{ou}\: \\ $$$$\mathrm{semi}−\mathrm{convergentes} \\ $$$$ \\ $$$$\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{n}} {\sum}}\frac{\left(-\mathrm{1}\right)^{\mathrm{k}-\mathrm{1}} }{\left(\mathrm{k}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$ \\ $$$$\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{n}} {\sum}}\frac{\left(-\mathrm{1}\right)^{\mathrm{k}-\mathrm{1}} }{\mathrm{k}} \\ $$$$ \\ $$$$\mathrm{Etudier}\:\mathrm{la}\:\mathrm{convergence}\:\mathrm{des}\:\mathrm{series}\:\mathrm{suivantes} \\ $$$$\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\frac{\left(\mathrm{2k}\right)!}{\mathrm{2}^{\mathrm{k}} } \\ $$$$\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{k}} \\ $$$$ \\ $$

Question Number 118950 Answers: 1 Comments: 0

lim_(x→∞) (1+(1/x))^x^2 .e^(−x) = ?.

$$\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{{x}^{\mathrm{2}} } .{e}^{−{x}} \:=\:?. \\ $$

Question Number 118949 Answers: 0 Comments: 0

find ∫_0 ^∞ ((sin(3cosx))/((x^2 +4)^2 ))dx

$${find}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{sin}\left(\mathrm{3}{cosx}\right)}{\left({x}^{\mathrm{2}} +\mathrm{4}\right)^{\mathrm{2}} }{dx} \\ $$

Question Number 118944 Answers: 1 Comments: 0

Question Number 118934 Answers: 0 Comments: 0

I=∫_0 ^∞ ((lnx)/(x^2 +2x+4)) dx put x+1=(√3) tanθ I=∫_0 ^(π/2) ((ln((√3) tanθ−1))/(3(sec^2 θ))) (√3) sec^2 θdθ I=(1/( (√3)))∫_0 ^(π/2) ln((√3)tanθ−1) dθ I=(1/( (√3)))∫_0 ^(π/2) ln((√3)sinθ−cosθ)−lncosθ dθ I=(1/( (√3)))∫_0 ^(π/2) ln(sin(θ−(π/6))+ln2−lncosθ dθ A=∫_(−2) ^6 ∣g(x)∣ dx ⇒let g(p)=0⇒ f(0)=p=2 A=∫_(−2) ^2 −g(x) dx+∫_2 ^6 g(x) dx ⇒put x=f(t) A=∫_(−1) ^0 −tf′(t) dt+∫_0 ^1 t f(t) dt A=∫_(−1) ^0 −3t^3 −3t dt+∫_0 ^1 3t^3 +3t dt A=2(((3t^4 )/4)+((3t^2 )/2))_0 ^1 =(9/2) lnx=(1/x)⇒xlnx=1(let x=α) ∫_1 ^α (1/x)−lnx dx=∫_α ^a lnx−(1/x) dx ⇒lnx−xlnx+x]_1 ^α =xlnx−x−lnx]_α ^a ⇒lnα−1+α−1=alna−a−lna−1+α +lnα ⇒alna−a−lna=−1 ⇒alna−lna=a−1⇒a=e s(t)=(1/2)∣OA^→ ×OB^→ ∣ s(t)=(1/2)∣(2,2,1)×(t,1,t+1)∣ s(t)=(1/2)∣(2t+1),(−2−t),(2−2t)∣ f(x)=(1/4)∫_0 ^x (2t+1)^2 +(t+2)^2 +(2−2t)^2 dt f(x)=(1/4)∫_0 ^x 9t^2 +9 dt=((3x^3 +9x)/4) A=(1/4)∫_0 ^6 (3x^3 +9x)dx=((3x^4 +18x^2 )/(16))]_0 ^6 A=((3.6^3 (6+1))/(16))=((567)/2)

$$ \\ $$$$ \\ $$$${I}=\int_{\mathrm{0}} ^{\infty} \:\frac{{lnx}}{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{4}}\:{dx} \\ $$$${put}\:{x}+\mathrm{1}=\sqrt{\mathrm{3}}\:{tan}\theta \\ $$$${I}=\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \:\frac{{ln}\left(\sqrt{\mathrm{3}}\:{tan}\theta−\mathrm{1}\right)}{\mathrm{3}\left({sec}^{\mathrm{2}} \theta\right)}\:\sqrt{\mathrm{3}}\:{sec}^{\mathrm{2}} \theta{d}\theta \\ $$$${I}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \:\:{ln}\left(\sqrt{\mathrm{3}}{tan}\theta−\mathrm{1}\right)\:{d}\theta \\ $$$${I}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\int_{\mathrm{0}} ^{\pi/\mathrm{2}} {ln}\left(\sqrt{\mathrm{3}}{sin}\theta−{cos}\theta\right)−{lncos}\theta\:{d}\theta \\ $$$${I}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\int_{\mathrm{0}} ^{\pi/\mathrm{2}} {ln}\left({sin}\left(\theta−\frac{\pi}{\mathrm{6}}\right)+{ln}\mathrm{2}−{lncos}\theta\:{d}\theta\right. \\ $$$$ \\ $$$$ \\ $$$${A}=\int_{−\mathrm{2}} ^{\mathrm{6}} \mid{g}\left({x}\right)\mid\:{dx} \\ $$$$\Rightarrow{let}\:{g}\left({p}\right)=\mathrm{0}\Rightarrow\:{f}\left(\mathrm{0}\right)={p}=\mathrm{2} \\ $$$${A}=\int_{−\mathrm{2}} ^{\mathrm{2}} −{g}\left({x}\right)\:{dx}+\int_{\mathrm{2}} ^{\mathrm{6}} {g}\left({x}\right)\:{dx} \\ $$$$\Rightarrow{put}\:{x}={f}\left({t}\right) \\ $$$${A}=\int_{−\mathrm{1}} ^{\mathrm{0}} −{tf}'\left({t}\right)\:{dt}+\int_{\mathrm{0}} ^{\mathrm{1}} {t}\:{f}\left({t}\right)\:{dt} \\ $$$${A}=\int_{−\mathrm{1}} ^{\mathrm{0}} −\mathrm{3}{t}^{\mathrm{3}} −\mathrm{3}{t}\:{dt}+\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{3}{t}^{\mathrm{3}} +\mathrm{3}{t}\:{dt} \\ $$$${A}=\mathrm{2}\left(\frac{\mathrm{3}{t}^{\mathrm{4}} }{\mathrm{4}}+\frac{\mathrm{3}{t}^{\mathrm{2}} }{\mathrm{2}}\right)_{\mathrm{0}} ^{\mathrm{1}} =\frac{\mathrm{9}}{\mathrm{2}} \\ $$$${lnx}=\frac{\mathrm{1}}{{x}}\Rightarrow{xlnx}=\mathrm{1}\left({let}\:{x}=\alpha\right) \\ $$$$\int_{\mathrm{1}} ^{\alpha} \frac{\mathrm{1}}{{x}}−{lnx}\:{dx}=\int_{\alpha} ^{{a}} {lnx}−\frac{\mathrm{1}}{{x}}\:{dx} \\ $$$$\left.\Rightarrow\left.{lnx}−{xlnx}+{x}\right]_{\mathrm{1}} ^{\alpha} ={xlnx}−{x}−{lnx}\right]_{\alpha} ^{{a}} \\ $$$$\Rightarrow{ln}\alpha−\mathrm{1}+\alpha−\mathrm{1}={alna}−{a}−{lna}−\mathrm{1}+\alpha \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+{ln}\alpha \\ $$$$\Rightarrow{alna}−{a}−{lna}=−\mathrm{1} \\ $$$$\Rightarrow{alna}−{lna}={a}−\mathrm{1}\Rightarrow{a}={e} \\ $$$$ \\ $$$$ \\ $$$${s}\left({t}\right)=\frac{\mathrm{1}}{\mathrm{2}}\mid{O}\overset{\rightarrow} {{A}}×{O}\overset{\rightarrow} {{B}}\mid \\ $$$${s}\left({t}\right)=\frac{\mathrm{1}}{\mathrm{2}}\mid\left(\mathrm{2},\mathrm{2},\mathrm{1}\right)×\left({t},\mathrm{1},{t}+\mathrm{1}\right)\mid \\ $$$${s}\left({t}\right)=\frac{\mathrm{1}}{\mathrm{2}}\mid\left(\mathrm{2}{t}+\mathrm{1}\right),\left(−\mathrm{2}−{t}\right),\left(\mathrm{2}−\mathrm{2}{t}\right)\mid \\ $$$${f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{{x}} \left(\mathrm{2}{t}+\mathrm{1}\right)^{\mathrm{2}} +\left({t}+\mathrm{2}\right)^{\mathrm{2}} +\left(\mathrm{2}−\mathrm{2}{t}\right)^{\mathrm{2}} \:{dt} \\ $$$${f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{{x}} \mathrm{9}{t}^{\mathrm{2}} +\mathrm{9}\:{dt}=\frac{\mathrm{3}{x}^{\mathrm{3}} +\mathrm{9}{x}}{\mathrm{4}} \\ $$$$\left.{A}=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\mathrm{6}} \left(\mathrm{3}{x}^{\mathrm{3}} +\mathrm{9}{x}\right){dx}=\frac{\mathrm{3}{x}^{\mathrm{4}} +\mathrm{18}{x}^{\mathrm{2}} }{\mathrm{16}}\right]_{\mathrm{0}} ^{\mathrm{6}} \\ $$$${A}=\frac{\mathrm{3}.\mathrm{6}^{\mathrm{3}} \left(\mathrm{6}+\mathrm{1}\right)}{\mathrm{16}}=\frac{\mathrm{567}}{\mathrm{2}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Question Number 118937 Answers: 1 Comments: 0

What is the general rule for factorising (a+b+c)^n ?

$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{general}\:\mathrm{rule}\: \\ $$$$\mathrm{for}\:\mathrm{factorising}\:\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)^{\mathrm{n}} ? \\ $$

Question Number 118935 Answers: 0 Comments: 0

Question Number 118930 Answers: 1 Comments: 0

Let [x] denote the greatest integer ≤x. Then the number of ordered pair (x,y), where x and y are positive integers less than 30 such that [(x/2)]+[((2x)/3)]+[(y/4)]+[((4y)/5)]=((7x)/6)+((21y)/(20)) is (A) 1 (B) 2 (C) 3 (D) 4

$$\mathrm{Let}\:\left[{x}\right]\:\mathrm{denote}\:\mathrm{the}\:\mathrm{greatest}\:\mathrm{integer}\:\leqslant{x}.\:\mathrm{Then}\:\mathrm{the} \\ $$$$\mathrm{number}\:\mathrm{of}\:\mathrm{ordered}\:\mathrm{pair}\:\left({x},\mathrm{y}\right),\:\mathrm{where}\:{x}\:\mathrm{and}\:\mathrm{y}\:\mathrm{are} \\ $$$$\mathrm{positive}\:\mathrm{integers}\:\mathrm{less}\:\mathrm{than}\:\mathrm{30}\:\mathrm{such}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\left[\frac{{x}}{\mathrm{2}}\right]+\left[\frac{\mathrm{2}{x}}{\mathrm{3}}\right]+\left[\frac{\mathrm{y}}{\mathrm{4}}\right]+\left[\frac{\mathrm{4y}}{\mathrm{5}}\right]=\frac{\mathrm{7}{x}}{\mathrm{6}}+\frac{\mathrm{21y}}{\mathrm{20}} \\ $$$$\mathrm{is} \\ $$$$\left(\mathrm{A}\right)\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\:\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{C}\right)\:\mathrm{3}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:\mathrm{4} \\ $$

Question Number 118928 Answers: 1 Comments: 2

Question Number 118927 Answers: 0 Comments: 0

Question Number 118924 Answers: 1 Comments: 0

... advanced calculus... prove that :: Σ_(n=1) ^∞ (((−1)^n H_n )/n^2 ) =∫_0 ^( 1) ((ln(1−x)ln(1+x) )/x)dx note :: H_n =Σ_(k=1) ^n (1/k) therefore: Σ_(n=1 ) ^∞ (((−1)^n H_n )/n^2 )=((−5)/8) ζ (3 ) ✓ ..m.n.july.1970...

$$\:\:\:\:\:\:\:\:\:...\:{advanced}\:{calculus}... \\ $$$$\:\:\:\:{prove}\:{that}\::: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} {H}_{{n}} }{{n}^{\mathrm{2}} }\:=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left(\mathrm{1}−{x}\right){ln}\left(\mathrm{1}+{x}\right)\:\:}{{x}}{dx}\:\: \\ $$$$\:\:\:\:\:{note}\:::\:{H}_{{n}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{k}} \\ $$$$\:\:\:\:\:\:\:{therefore}:\:\underset{{n}=\mathrm{1}\:} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} {H}_{{n}} }{{n}^{\mathrm{2}} }=\frac{−\mathrm{5}}{\mathrm{8}}\:\zeta\:\left(\mathrm{3}\:\right)\:\checkmark \\ $$$$\:\:\:\:\:\:\:\:..{m}.{n}.{july}.\mathrm{1970}... \\ $$$$ \\ $$

Question Number 118917 Answers: 0 Comments: 1

x(x+y)dy−x^2 dx

$${x}\left({x}+{y}\right){dy}−{x}^{\mathrm{2}} {dx} \\ $$

Question Number 118936 Answers: 3 Comments: 0

show by recurrence that: for n ∈ N^∗ , 2^(6n−5) +3^(2n ) is divisible by 11.

$${show}\:{by}\:{recurrence}\:{that}: \\ $$$${for}\:{n}\:\in\:\mathbb{N}^{\ast} ,\:\mathrm{2}^{\mathrm{6}{n}−\mathrm{5}} +\mathrm{3}^{\mathrm{2}{n}\:} \:{is}\:{divisible}\:{by} \\ $$$$\mathrm{11}. \\ $$

Question Number 118907 Answers: 0 Comments: 0

where can u fond a formula for denesting ramanujan type roots?

$${where}\:{can}\:{u}\:{fond}\:{a}\:{formula}\:{for} \\ $$$${denesting}\:{ramanujan}\:{type}\:{roots}? \\ $$

Question Number 118905 Answers: 4 Comments: 0

∫ (dλ/((λ^2 −9)^2 )) =?