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Question Number 111957    Answers: 0   Comments: 31

“MATHEMATICS” CONTAINS ALL THE LETTERS OF “ETHICS”. IS THERE ANY LESSON FOR US IN ABOVE SAYING? FOR “math-lovers”? FOR “math-giants”? FOR “overflow-mathematicians”? ........ ...... _( BTW this saying belongs to me)

$$\:\:\:\:\:\:\:\:\:\:``\mathrm{MATHEMATICS}'' \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{CONTAINS} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{ALL}\:\mathrm{THE}\:\mathrm{LETTERS}\:\mathrm{OF} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:``\mathrm{ETHICS}''. \\ $$$$ \\ $$$$\mathrm{IS}\:\:\mathrm{THERE}\:\mathrm{ANY}\:\mathrm{LESSON}\:\mathrm{FOR}\:\mathrm{US} \\ $$$$\mathrm{IN}\:\mathrm{ABOVE}\:\mathrm{SAYING}? \\ $$$${FOR}\:``{math}-{lovers}''? \\ $$$${FOR}\:``{math}-{giants}''? \\ $$$${FOR}\:``{overflow}-{mathematicians}''? \\ $$$$........ \\ $$$$......\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:_{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{BTW}\:{this}\:{saying}\:{belongs}\:{to}\:{me}} \\ $$

Question Number 111954    Answers: 1   Comments: 0

Proof for the following : n! ≥ 2×3^(n−2)

$${P}\mathrm{roof}\:\mathrm{for}\:\mathrm{the}\:\mathrm{following}\:: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:{n}!\:\geqslant\:\mathrm{2}×\mathrm{3}^{{n}−\mathrm{2}} \\ $$$$ \\ $$

Question Number 111953    Answers: 0   Comments: 0

Question Number 112004    Answers: 1   Comments: 0

Suppose x,y,z ∈N, (yz+x) is prime (yz+x)∣(zx+y), (yz+x)∣(xy+z). Find all possible values of (((xy+z)(zx+y))/((yz+x)^2 )).

$$\mathrm{Suppose}\:\mathrm{x},\mathrm{y},\mathrm{z}\:\in\mathbb{N},\:\left(\mathrm{yz}+\mathrm{x}\right)\:\mathrm{is}\:\mathrm{prime} \\ $$$$\left(\mathrm{yz}+\mathrm{x}\right)\mid\left(\mathrm{zx}+\mathrm{y}\right),\:\left(\mathrm{yz}+\mathrm{x}\right)\mid\left(\mathrm{xy}+\mathrm{z}\right). \\ $$$$\mathrm{Find}\:\mathrm{all}\:\mathrm{possible}\:\mathrm{values}\:\mathrm{of}\: \\ $$$$\frac{\left(\mathrm{xy}+\mathrm{z}\right)\left(\mathrm{zx}+\mathrm{y}\right)}{\left(\mathrm{yz}+\mathrm{x}\right)^{\mathrm{2}} }. \\ $$

Question Number 112545    Answers: 2   Comments: 2

please solve : I=∫_0 ^( 1) xlog^2 (((1−x)/(1+x)))dx =??? ...m.n.july 1970.... good luck .

$$\:\:\:\:{please}\:{solve}\:: \\ $$$$ \\ $$$$\:\:\:\:\:\mathrm{I}=\int_{\mathrm{0}} ^{\:\mathrm{1}} {xlog}^{\mathrm{2}} \left(\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\right){dx}\:=??? \\ $$$$ \\ $$$$\:\:\:\:\:\:...{m}.{n}.{july}\:\mathrm{1970}.... \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{good}\:\:\:{luck}\:. \\ $$$$ \\ $$

Question Number 111939    Answers: 2   Comments: 0

Question Number 111937    Answers: 1   Comments: 0

Question Number 112203    Answers: 0   Comments: 2

proporsed by m.njuly 1970 ∫_0 ^1 ((ln(((x+1))^(1/3) ))/((x+1)(x+2)))dx solution let I=(1/3)∫_0 ^1 ((ln(x+1))/((x+1)(x+2)))dx I=(1/3)∫_0 ^1 ((ln(1+x))/(x+1))dx−∫_0 ^1 ((ln(x+1))/(x+2))dx=A−B let A=(1/3)∫_0 ^1 ((ln(x+1))/(x+1))dx ( using IBP) A=(1/3)[ln^2 (x+1)]_0 ^1 −(1/3)∫_0 ^1 ((ln(x+1))/(x+1))dx A(1+(1/3))=(1/3)ln^2 (2) A=∫_0 ^1 ((ln(x+1))/(x+1))dx=(1/3)ln^2 (2)......(1) then B=(1/3)∫_0 ^1 ((ln(x+1))/(x+2))dx B=(1/3)[ln(x+2)ln(x+1)]_0 ^1 −(1/3)∫_0 ^1 ((ln(x+2))/(x+1))dx let x+1=t B=(1/3)ln3ln 2−(1/3)∫_1 ^2 ((ln(1+t))/t)dt let t=−x B=(1/3)ln3ln2−(1/3)∫_(−1) ^(−2) ((ln(1−x))/x)dx but ∫((ln(1−x))/x)dx=−Li_2 (x) B=(1/3)ln3ln2+(1/3)[Li_2 (x)]_(−1) ^(−2) B=(1/3)ln3ln2+(1/3)Li_2 (−2)−(1/3)Li_2 (−1) B=(1/3)ln3ln2+(1/3)Li_2 (−2)+(π^2 /(36)) but I=A−B ∫_0 ^1 ((ln(((1+x))^(1/3) ))/((x+1)(x+2)))dx=(1/4)ln^2 (2)−(1/3)ln3ln2−(1/3)Li_2 (−2)−(π^2 /(36)) mathdave(06/09/2020)

$${proporsed}\:{by}\:{m}.{njuly}\:\mathrm{1970} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\sqrt[{\mathrm{3}}]{{x}+\mathrm{1}}\right)}{\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)}{dx} \\ $$$${solution} \\ $$$${let}\:{I}=\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left({x}+\mathrm{1}\right)}{\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)}{dx} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}+{x}\right)}{{x}+\mathrm{1}}{dx}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left({x}+\mathrm{1}\right)}{{x}+\mathrm{2}}{dx}={A}−{B} \\ $$$${let}\:{A}=\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left({x}+\mathrm{1}\right)}{{x}+\mathrm{1}}{dx}\:\left(\:{using}\:{IBP}\right) \\ $$$${A}=\frac{\mathrm{1}}{\mathrm{3}}\left[\mathrm{ln}^{\mathrm{2}} \left({x}+\mathrm{1}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} −\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left({x}+\mathrm{1}\right)}{{x}+\mathrm{1}}{dx} \\ $$$${A}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}\right)=\frac{\mathrm{1}}{\mathrm{3}}\mathrm{ln}^{\mathrm{2}} \left(\mathrm{2}\right) \\ $$$${A}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left({x}+\mathrm{1}\right)}{{x}+\mathrm{1}}{dx}=\frac{\mathrm{1}}{\mathrm{3}}\mathrm{ln}^{\mathrm{2}} \left(\mathrm{2}\right)......\left(\mathrm{1}\right) \\ $$$${then}\:{B}=\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left({x}+\mathrm{1}\right)}{{x}+\mathrm{2}}{dx} \\ $$$${B}=\frac{\mathrm{1}}{\mathrm{3}}\left[\mathrm{ln}\left({x}+\mathrm{2}\right)\mathrm{ln}\left({x}+\mathrm{1}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} −\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left({x}+\mathrm{2}\right)}{{x}+\mathrm{1}}{dx} \\ $$$${let}\:{x}+\mathrm{1}={t} \\ $$$${B}=\frac{\mathrm{1}}{\mathrm{3}}\mathrm{ln3ln}\:\mathrm{2}−\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{1}} ^{\mathrm{2}} \frac{\mathrm{ln}\left(\mathrm{1}+{t}\right)}{{t}}{dt} \\ $$$${let}\:{t}=−{x} \\ $$$${B}=\frac{\mathrm{1}}{\mathrm{3}}\mathrm{ln3ln2}−\frac{\mathrm{1}}{\mathrm{3}}\int_{−\mathrm{1}} ^{−\mathrm{2}} \frac{\mathrm{ln}\left(\mathrm{1}−{x}\right)}{{x}}{dx} \\ $$$${but}\:\int\frac{\mathrm{ln}\left(\mathrm{1}−{x}\right)}{{x}}{dx}=−{Li}_{\mathrm{2}} \left({x}\right) \\ $$$${B}=\frac{\mathrm{1}}{\mathrm{3}}\mathrm{ln3ln2}+\frac{\mathrm{1}}{\mathrm{3}}\left[{Li}_{\mathrm{2}} \left({x}\right)\right]_{−\mathrm{1}} ^{−\mathrm{2}} \\ $$$${B}=\frac{\mathrm{1}}{\mathrm{3}}\mathrm{ln3ln2}+\frac{\mathrm{1}}{\mathrm{3}}{Li}_{\mathrm{2}} \left(−\mathrm{2}\right)−\frac{\mathrm{1}}{\mathrm{3}}{Li}_{\mathrm{2}} \left(−\mathrm{1}\right) \\ $$$${B}=\frac{\mathrm{1}}{\mathrm{3}}\mathrm{ln3ln2}+\frac{\mathrm{1}}{\mathrm{3}}{Li}_{\mathrm{2}} \left(−\mathrm{2}\right)+\frac{\pi^{\mathrm{2}} }{\mathrm{36}} \\ $$$${but}\:{I}={A}−{B} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\sqrt[{\mathrm{3}}]{\mathrm{1}+{x}}\right)}{\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)}{dx}=\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}^{\mathrm{2}} \left(\mathrm{2}\right)−\frac{\mathrm{1}}{\mathrm{3}}\mathrm{ln3ln2}−\frac{\mathrm{1}}{\mathrm{3}}{Li}_{\mathrm{2}} \left(−\mathrm{2}\right)−\frac{\pi^{\mathrm{2}} }{\mathrm{36}} \\ $$$${mathdave}\left(\mathrm{06}/\mathrm{09}/\mathrm{2020}\right) \\ $$

Question Number 111934    Answers: 1   Comments: 0

Question Number 112533    Answers: 2   Comments: 2

Find the minimum number of n integers to be selected from S={1,2,3,...11} so that the difference of two of the n integers is 7.

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{number}\:\mathrm{of}\:\mathrm{n} \\ $$$$\mathrm{integers}\:\mathrm{to}\:\mathrm{be}\:\mathrm{selected}\:\mathrm{from} \\ $$$$\mathrm{S}=\left\{\mathrm{1},\mathrm{2},\mathrm{3},...\mathrm{11}\right\}\:\mathrm{so}\:\mathrm{that}\:\mathrm{the}\:\mathrm{difference} \\ $$$$\mathrm{of}\:\mathrm{two}\:\mathrm{of}\:\mathrm{the}\:\mathrm{n}\:\mathrm{integers}\:\mathrm{is}\:\mathrm{7}. \\ $$

Question Number 112060    Answers: 1   Comments: 0

In a trapezium, ABCD, with AB parallel to CD. If M is the midpoint of line segment AD and P is a point on line BC such that MP is perpendicular to BC. Show that, we need only the lengths of line segments MP and BC to calculate the area ABCD.

$$\mathrm{In}\:\mathrm{a}\:\mathrm{trapezium},\:\mathrm{ABCD},\:\mathrm{with}\:\mathrm{AB} \\ $$$$\mathrm{parallel}\:\mathrm{to}\:\mathrm{CD}.\:\mathrm{If}\:\mathrm{M}\:\mathrm{is}\:\mathrm{the}\:\mathrm{midpoint}\:\mathrm{of} \\ $$$$\mathrm{line}\:\mathrm{segment}\:\mathrm{AD}\:\mathrm{and}\:\mathrm{P}\:\mathrm{is}\:\mathrm{a}\:\mathrm{point}\:\mathrm{on} \\ $$$$\mathrm{line}\:\mathrm{BC}\:\mathrm{such}\:\mathrm{that}\:\mathrm{MP}\:\mathrm{is}\:\mathrm{perpendicular} \\ $$$$\mathrm{to}\:\mathrm{BC}.\:\mathrm{Show}\:\mathrm{that},\:\mathrm{we}\:\mathrm{need}\:\mathrm{only}\:\mathrm{the} \\ $$$$\mathrm{lengths}\:\mathrm{of}\:\mathrm{line}\:\mathrm{segments}\:\mathrm{MP}\:\mathrm{and}\:\mathrm{BC} \\ $$$$\mathrm{to}\:\mathrm{calculate}\:\mathrm{the}\:\mathrm{area}\:\mathrm{ABCD}. \\ $$

Question Number 111926    Answers: 1   Comments: 1

Question Number 112001    Answers: 0   Comments: 2

Question Number 111923    Answers: 2   Comments: 0

(√(bemath )) lim_(x→0) ((((cos 4x))^(1/(3 )) −1)/(cos 3x−cos 9x)) ?

$$\sqrt{{bemath}\:} \\ $$$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{3}\:}]{\mathrm{cos}\:\mathrm{4}{x}}\:−\mathrm{1}}{\mathrm{cos}\:\mathrm{3}{x}−\mathrm{cos}\:\mathrm{9}{x}}\:? \\ $$

Question Number 111982    Answers: 2   Comments: 0

Question Number 111914    Answers: 1   Comments: 0

lim_(x→0) ((8x^3 )/( (√(4+sin 6x)) −(√(tan 6x+4)))) ?

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{8}{x}^{\mathrm{3}} }{\:\sqrt{\mathrm{4}+\mathrm{sin}\:\mathrm{6}{x}}\:−\sqrt{\mathrm{tan}\:\mathrm{6}{x}+\mathrm{4}}}\:? \\ $$

Question Number 111909    Answers: 2   Comments: 0

solve { ((x(√x)+y(√y) = 5)),((x(√y) +y(√x) = 1)) :}

$${solve}\:\begin{cases}{{x}\sqrt{{x}}+{y}\sqrt{{y}}\:=\:\mathrm{5}}\\{{x}\sqrt{{y}}\:+{y}\sqrt{{x}}\:=\:\mathrm{1}}\end{cases} \\ $$

Question Number 111907    Answers: 2   Comments: 0

lim_(x→(π/(20))) ((1−tan 5x)/(sin 5x−cos 5x)) ?

$$\underset{{x}\rightarrow\frac{\pi}{\mathrm{20}}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{tan}\:\mathrm{5}{x}}{\mathrm{sin}\:\mathrm{5}{x}−\mathrm{cos}\:\mathrm{5}{x}}\:? \\ $$

Question Number 111906    Answers: 1   Comments: 0

Question Number 111890    Answers: 0   Comments: 0

Question Number 111879    Answers: 1   Comments: 0

prove that lim_(n→∞) nΣ_(k=1) ^(n−1) ((ln(k+n)−ln(n))/(k^2 +n^2 ))=(π/8)ln2

$${prove}\:{that}\: \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}{n}\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}\frac{\mathrm{ln}\left({k}+{n}\right)−\mathrm{ln}\left({n}\right)}{{k}^{\mathrm{2}} +{n}^{\mathrm{2}} }=\frac{\pi}{\mathrm{8}}\mathrm{ln2} \\ $$

Question Number 111876    Answers: 0   Comments: 0

Question Number 111873    Answers: 1   Comments: 0

(√(bemath)) ∫ ((2−cos x)/(2+cos x)) dx

$$\:\:\:\sqrt{{bemath}} \\ $$$$\int\:\frac{\mathrm{2}−\mathrm{cos}\:{x}}{\mathrm{2}+\mathrm{cos}\:{x}}\:{dx}\: \\ $$

Question Number 111869    Answers: 0   Comments: 0

evaluate ∫_0 ^1 ((ln(1+x))/(x^2 +5x+6))dx

$${evaluate} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}+{x}\right)}{{x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{6}}{dx} \\ $$

Question Number 111868    Answers: 2   Comments: 0

simplify ∫((1−5sin^2 x)/(cos^5 xsin^2 x))dx

$${simplify} \\ $$$$\int\frac{\mathrm{1}−\mathrm{5sin}^{\mathrm{2}} {x}}{\mathrm{cos}^{\mathrm{5}} {x}\mathrm{sin}^{\mathrm{2}} {x}}{dx} \\ $$

Question Number 111859    Answers: 2   Comments: 7

I=∫(dx/((x^2 +2x+3)(√(x^2 +x+3)))) = ? my try..

$${I}=\int\frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{3}\right)\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{3}}}\:=\:? \\ $$$${my}\:{try}.. \\ $$

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