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Question Number 141189    Answers: 0   Comments: 3

find the area of the shaded region shown below which is boinded by to functions f(x)=x^2 g(x)=2−x and the x-axis

$${find}\:{the}\:{area}\:{of}\:{the}\:{shaded}\:{region} \\ $$$${shown}\:{below}\:{which}\:{is}\:{boinded}\:{by}\:{to}\:{functions}\: \\ $$$${f}\left({x}\right)={x}^{\mathrm{2}} \:\:{g}\left({x}\right)=\mathrm{2}−{x}\:{and}\:{the}\:{x}-{axis} \\ $$$$ \\ $$

Question Number 141142    Answers: 0   Comments: 0

Question Number 141134    Answers: 1   Comments: 0

prove that:: Φ:=Σ_(m,n=1) ^∞ (((−1)^(n−1) )/(m^2 +n^2 ))= (π^2 /(24))+((πln(2))/8)

$$\:\:\:\:\: \\ $$$$\:\:\:{prove}\:{that}:: \\ $$$$\:\:\Phi:=\underset{{m},{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{m}^{\mathrm{2}} +{n}^{\mathrm{2}} }=\:\frac{\pi^{\mathrm{2}} }{\mathrm{24}}+\frac{\pi{ln}\left(\mathrm{2}\right)}{\mathrm{8}} \\ $$$$ \\ $$

Question Number 141133    Answers: 1   Comments: 0

I:=∫_0 ^( ∞) (((x^n −1)(x−1))/(x^(n+3) −1))dx=??

$$\: \\ $$$$\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{I}}:=\int_{\mathrm{0}} ^{\:\infty} \frac{\left({x}^{{n}} −\mathrm{1}\right)\left({x}−\mathrm{1}\right)}{{x}^{{n}+\mathrm{3}} −\mathrm{1}}{dx}=?? \\ $$$$ \\ $$

Question Number 141085    Answers: 2   Comments: 0

Question Number 141082    Answers: 3   Comments: 0

........ nice ....... calculus ........ 𝛗:=Σ_(n=2) ^∞ ((ζ ( n ))/(n . 4^n ))=?

$$\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:........\:{nice}\:\:.......\:\:{calculus}\:........ \\ $$$$\:\:\:\boldsymbol{\phi}:=\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\:\:\frac{\zeta\:\left(\:{n}\:\right)}{{n}\:.\:\mathrm{4}^{{n}} }=? \\ $$$$ \\ $$

Question Number 141031    Answers: 0   Comments: 0

∫_(π/6) ^( π/3) (√(1+((cos^2 x)/(sin x)))) dx ?

$$\:\:\:\:\:\:\int_{\pi/\mathrm{6}} ^{\:\pi/\mathrm{3}} \:\sqrt{\mathrm{1}+\frac{\mathrm{cos}\:^{\mathrm{2}} {x}}{\mathrm{sin}\:{x}}}\:{dx}\:?\: \\ $$

Question Number 140996    Answers: 1   Comments: 0

Evaluation of :: Ω :=Σ_(n=0) ^∞ (((−1)^n )/(1+n^2 )) solution:: Ω:=1+Σ_(n=1) ^∞ (((−1)^n )/(n^2 −i^2 )) =(1/(2i)){Σ_(n=1) ^∞ (((−1)^n )/(n−i))−(((−1)^n )/(n+i))} :=1+(1/(2i)) (Φ−Ψ) where Φ:=Σ_(n=1) ^∞ (((−1)^n )/(n−i)) and Ψ :=Σ_(n=1) ^∞ (((−1)^n )/(n+i)) Φ:=Σ_(n=1) ^∞ (1/(2n−k)) −Σ_(n=1) ^∞ (1/(2n−1−i)) :=(1/2){Σ_(n=1) ^∞ (1/(n−(i/2)))−Σ_(n=1) ^∞ (1/(n−((1+i)/2)))} :=(1/2){ψ(1−((1+i)/2))−ψ(1−(i/2))} :=(1/2)(ψ(((1−i)/2))−ψ(1−(i/2))).... Ψ:=Σ_(n=1) ^∞ (((−1)^n )/(n+i)) =Σ_(n=1) ^∞ (1/(2n+i))−Σ_(n=1) ^∞ (1/(2n−1+i)) :=(1/2){Σ_(n=1) ^∞ (1/(n+(i/2)))−Σ_(n=1) ^∞ (1/(n+((i−1)/2)))} :=(1/2)(ψ(((1+i)/2))−ψ(1+(i/2))) .... Φ−Ψ:=(1/2){ψ(((1−i)/2))−ψ(((1+i)/2))} +(1/2){ψ(1+(i/2))−ψ(1−(i/2))} :=(1/2)(−πcotπ(((1−i)/2)))+(1/2)((2/i)−πcot(π(i/2))) :=−(π/2)tan(((πi)/2))−(π/2)cot(((πi)/2))−i :=−i−(π/(sin(πi)))=−i−((2iπ)/(e^(−π) −e^π )) :=−i+πicsch(π) .... Ω :=1+(1/(2i))(−i+πicsch(π))=(1/2)+(π/2) csch(π) ... Ω:=(1/2)+(π/2) csch(π)....✓✓✓

$$ \\ $$$$\:\:\:\:\:\:\mathscr{E}{valuation}\:{of}\:::\:\Omega\::=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{1}+{n}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:{solution}:: \\ $$$$\:\:\:\:\:\Omega:=\mathrm{1}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} −{i}^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\mathrm{2}{i}}\left\{\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}−{i}}−\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+{i}}\right\} \\ $$$$\:\:\:\:\:\:\::=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}{i}}\:\left(\Phi−\Psi\right)\:\:\:\:{where}\:\:\Phi:=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}−{i}} \\ $$$$\:\:\:\:\:\:\:{and}\:\:\:\:\Psi\::=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+{i}}\:\:\: \\ $$$$\:\:\:\:\:\:\:\Phi:=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}{n}−{k}}\:−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}−{i}} \\ $$$$\:\:\:\:\:\:\:\::=\frac{\mathrm{1}}{\mathrm{2}}\left\{\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}−\frac{{i}}{\mathrm{2}}}−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}−\frac{\mathrm{1}+{i}}{\mathrm{2}}}\right\} \\ $$$$\:\:\:\:\:\:\:\::=\frac{\mathrm{1}}{\mathrm{2}}\left\{\psi\left(\mathrm{1}−\frac{\mathrm{1}+{i}}{\mathrm{2}}\right)−\psi\left(\mathrm{1}−\frac{{i}}{\mathrm{2}}\right)\right\} \\ $$$$\:\:\:\:\:\:\:\::=\frac{\mathrm{1}}{\mathrm{2}}\left(\psi\left(\frac{\mathrm{1}−{i}}{\mathrm{2}}\right)−\psi\left(\mathrm{1}−\frac{{i}}{\mathrm{2}}\right)\right).... \\ $$$$\:\:\:\:\Psi:=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+{i}}\:=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}{n}+{i}}−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}+{i}} \\ $$$$\:\:\:\:\:\:\:\::=\frac{\mathrm{1}}{\mathrm{2}}\left\{\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}+\frac{{i}}{\mathrm{2}}}−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}+\frac{{i}−\mathrm{1}}{\mathrm{2}}}\right\} \\ $$$$\:\:\:\:\:\:\::=\frac{\mathrm{1}}{\mathrm{2}}\left(\psi\left(\frac{\mathrm{1}+{i}}{\mathrm{2}}\right)−\psi\left(\mathrm{1}+\frac{{i}}{\mathrm{2}}\right)\right)\:.... \\ $$$$\:\:\:\:\:\Phi−\Psi:=\frac{\mathrm{1}}{\mathrm{2}}\left\{\psi\left(\frac{\mathrm{1}−{i}}{\mathrm{2}}\right)−\psi\left(\frac{\mathrm{1}+{i}}{\mathrm{2}}\right)\right\} \\ $$$$\:\:\:\:\:\:\:\:\:+\frac{\mathrm{1}}{\mathrm{2}}\left\{\psi\left(\mathrm{1}+\frac{{i}}{\mathrm{2}}\right)−\psi\left(\mathrm{1}−\frac{{i}}{\mathrm{2}}\right)\right\} \\ $$$$\:\:\:\:\:\:\::=\frac{\mathrm{1}}{\mathrm{2}}\left(−\pi{cot}\pi\left(\frac{\mathrm{1}−{i}}{\mathrm{2}}\right)\right)+\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{2}}{{i}}−\pi{cot}\left(\pi\frac{{i}}{\mathrm{2}}\right)\right) \\ $$$$\:\:\:\:\::=−\frac{\pi}{\mathrm{2}}{tan}\left(\frac{\pi{i}}{\mathrm{2}}\right)−\frac{\pi}{\mathrm{2}}{cot}\left(\frac{\pi{i}}{\mathrm{2}}\right)−{i} \\ $$$$\:\:\:\::=−{i}−\frac{\pi}{{sin}\left(\pi{i}\right)}=−{i}−\frac{\mathrm{2}{i}\pi}{{e}^{−\pi} −{e}^{\pi} } \\ $$$$\:\:\:\::=−{i}+\pi{icsch}\left(\pi\right)\:.... \\ $$$$\:\:\:\:\:\:\Omega\::=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}{i}}\left(−{i}+\pi{icsch}\left(\pi\right)\right)=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\pi}{\mathrm{2}}\:{csch}\left(\pi\right)\:... \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\Omega:=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\pi}{\mathrm{2}}\:{csch}\left(\pi\right)....\checkmark\checkmark\checkmark \\ $$$$\:\:\:\:\:\:\:\: \\ $$

Question Number 140977    Answers: 2   Comments: 0

find ∫_0 ^∞ (e^(−t(1+x^2 )) /(1+x^2 ))dx with t≥0

$${find}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{e}^{−{t}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)} }{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:\:{with}\:{t}\geqslant\mathrm{0} \\ $$

Question Number 140976    Answers: 1   Comments: 0

find ∫_0 ^π (dx/((2−cosx−sinx)^2 ))

$${find}\:\int_{\mathrm{0}} ^{\pi} \:\frac{{dx}}{\left(\mathrm{2}−{cosx}−{sinx}\right)^{\mathrm{2}} } \\ $$$$ \\ $$

Question Number 140971    Answers: 0   Comments: 1

Question Number 140961    Answers: 2   Comments: 0

∫_0 ^( 1) ((ln (x+(√(1−x^2 ))))/x) dx =?

$$\:\underset{\mathrm{0}} {\overset{\:\mathrm{1}} {\int}}\:\:\frac{\mathrm{ln}\:\left({x}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)}{{x}}\:{dx}\:=?\: \\ $$

Question Number 140956    Answers: 1   Comments: 0

.....advanced......calculus..... prove that: 𝛗:= ∫_(−∞) ^( ∞) ((sin^4 (x).cos^4 (x))/x^2 )dx=(π/(16)) m.n

$$\:\:\:\:\:\:\:\:\:\:.....{advanced}......{calculus}..... \\ $$$$\:\:\:\:\:{prove}\:{that}: \\ $$$$\:\:\boldsymbol{\phi}:=\:\int_{−\infty} ^{\:\infty} \frac{{sin}^{\mathrm{4}} \left({x}\right).{cos}^{\mathrm{4}} \left({x}\right)}{{x}^{\mathrm{2}} }{dx}=\frac{\pi}{\mathrm{16}} \\ $$$$\:\:{m}.{n} \\ $$

Question Number 140966    Answers: 1   Comments: 0

.......nice......calculus..... if Σ_(n=0) ^∞ (((√(cos (nπ))) )/((2n)!!)) = ω then Re(ω):=??

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:.......{nice}......{calculus}..... \\ $$$$\:\:\:\:\:{if}\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\sqrt{{cos}\:\left({n}\pi\right)}\:}{\left(\mathrm{2}{n}\right)!!}\:=\:\omega \\ $$$$\:\:\:\:\:\:\:{then}\:\:\:{Re}\left(\omega\right):=?? \\ $$

Question Number 140930    Answers: 3   Comments: 0

∫_0 ^π (dx/( (√2) −cos x))

$$\underset{\mathrm{0}} {\overset{\pi} {\int}}\:\frac{\mathrm{dx}}{\:\sqrt{\mathrm{2}}\:−\mathrm{cos}\:\mathrm{x}} \\ $$

Question Number 140896    Answers: 2   Comments: 0

the function f with variable x satisfies the equation x^2 f ′(x) +2x f(x) = arctan x for 0 < arctan x <(π/2) and f(1)=(π/4). find f(x).

$$\mathrm{the}\:\mathrm{function}\:\mathrm{f}\:\mathrm{with}\:\mathrm{variable}\:\mathrm{x} \\ $$$$\mathrm{satisfies}\:\mathrm{the}\:\mathrm{equation}\: \\ $$$$\mathrm{x}^{\mathrm{2}} \:\mathrm{f}\:'\left(\mathrm{x}\right)\:+\mathrm{2x}\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\mathrm{arctan}\:\mathrm{x}\:\mathrm{for}\: \\ $$$$\mathrm{0}\:<\:\mathrm{arctan}\:\mathrm{x}\:<\frac{\pi}{\mathrm{2}}\:\mathrm{and}\:\mathrm{f}\left(\mathrm{1}\right)=\frac{\pi}{\mathrm{4}}. \\ $$$$\mathrm{find}\:\mathrm{f}\left(\mathrm{x}\right). \\ $$

Question Number 140866    Answers: 1   Comments: 0

∫_0 ^1 ((ln 2−ln (1+x^2 ))/(1−x)) dx =?

$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\frac{\mathrm{ln}\:\mathrm{2}−\mathrm{ln}\:\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{1}−\mathrm{x}}\:\mathrm{dx}\:=?\: \\ $$

Question Number 140885    Answers: 1   Comments: 0

....... Advanced ::::::::::★★★:::::::::: Calculus....... find the value of the infinite series:: Θ := Σ_(n=1) ^∞ (((−1)^(n−1) H_( 2n) )/(2n−1)) = ??? .......M.N.july.1970........

$$\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:.......\:{Advanced}\:::::::::::\bigstar\bigstar\bigstar::::::::::\:{Calculus}....... \\ $$$$\:\:\:\:\:\:{find}\:{the}\:{value}\:{of}\:\:{the}\:{infinite}\:{series}:: \\ $$$$\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\Theta\::=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \mathrm{H}_{\:\mathrm{2}{n}} }{\mathrm{2}{n}−\mathrm{1}}\:=\:??? \\ $$$$\:\:\:\:\:\:\:\:\:\:.......\mathscr{M}.\mathscr{N}.{july}.\mathrm{1970}........ \\ $$

Question Number 140823    Answers: 1   Comments: 1

Determine whether the improper integral converges or diverges ∫_1 ^( ∞) ((2x+7)/(7x^3 +5x^2 +1)) dx

$$\mathrm{Determine}\:\mathrm{whether}\:\mathrm{the}\:\mathrm{improper} \\ $$$$\mathrm{integral}\:\mathrm{converges}\:\mathrm{or}\:\mathrm{diverges}\: \\ $$$$\int_{\mathrm{1}} ^{\:\infty} \:\frac{\mathrm{2x}+\mathrm{7}}{\mathrm{7x}^{\mathrm{3}} +\mathrm{5x}^{\mathrm{2}} +\mathrm{1}}\:\mathrm{dx}\: \\ $$

Question Number 140793    Answers: 0   Comments: 0

∫(√((x+2)/e^x ))dx=...?

$$\int\sqrt{\frac{{x}+\mathrm{2}}{{e}^{{x}} }}{dx}=...? \\ $$

Question Number 140789    Answers: 2   Comments: 0

.......nice.....math.... calculate:: Θ:= Σ_(n=1) ^∞ (1/) =??

$$\:\:\:\:\:\:\:\:.......{nice}.....{math}.... \\ $$$$\:\:\:\:\:\:{calculate}::\:\Theta:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{}\:=?? \\ $$$$ \\ $$

Question Number 140768    Answers: 3   Comments: 0

∫_(−∞) ^∞ ((x^2 +4)/(x^4 +16)) dx =?

$$\:\int_{−\infty} ^{\infty} \frac{\mathrm{x}^{\mathrm{2}} +\mathrm{4}}{\mathrm{x}^{\mathrm{4}} +\mathrm{16}}\:\mathrm{dx}\:=? \\ $$

Question Number 140715    Answers: 1   Comments: 0

......advanced calculus...... prove that: 𝛗:=∫_0 ^( ∞) (x^2 /(cosh^2 (x^2 )))dx=^? (((√2) −2)/4) (√π) ζ ( (1/2) ) ..............

$$\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:......{advanced}\:\:{calculus}...... \\ $$$$\:\:\:\:\:\:{prove}\:{that}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\boldsymbol{\phi}:=\int_{\mathrm{0}} ^{\:\infty} \frac{{x}^{\mathrm{2}} }{{cosh}^{\mathrm{2}} \left({x}^{\mathrm{2}} \right)}{dx}\overset{?} {=}\frac{\sqrt{\mathrm{2}}\:−\mathrm{2}}{\mathrm{4}}\:\sqrt{\pi}\:\zeta\:\left(\:\frac{\mathrm{1}}{\mathrm{2}}\:\right)\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:.............. \\ $$

Question Number 140703    Answers: 2   Comments: 0

Question Number 140684    Answers: 2   Comments: 0

∫_0 ^π cos^n (x)∙cos (nx)dx=(π/2^n )

$$\int_{\mathrm{0}} ^{\pi} \mathrm{cos}\:^{{n}} \left({x}\right)\centerdot\mathrm{cos}\:\left({nx}\right){dx}=\frac{\pi}{\mathrm{2}^{{n}} } \\ $$

Question Number 140685    Answers: 0   Comments: 0

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