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IntegrationQuestion and Answers: Page 101

Question Number 135784 Answers: 1 Comments: 0

Given { ((f(3)=4 , f ′(3)=−2)),((f(8)=5 , f ′(8)=3)) :} find ∫_3 ^( 8) x f ′′(x) dx .

$${Given}\:\begin{cases}{{f}\left(\mathrm{3}\right)=\mathrm{4}\:,\:{f}\:'\left(\mathrm{3}\right)=−\mathrm{2}}\\{{f}\left(\mathrm{8}\right)=\mathrm{5}\:,\:{f}\:'\left(\mathrm{8}\right)=\mathrm{3}}\end{cases} \\ $$$${find}\:\int_{\mathrm{3}} ^{\:\mathrm{8}} \:{x}\:{f}\:''\left({x}\right)\:{dx}\:. \\ $$

Question Number 135753 Answers: 1 Comments: 1

Question Number 135777 Answers: 1 Comments: 0

let U_n =∫_(−∞) ^(+∞) ((cos(nx))/((x^2 −x+1)^2 ))dx calculate lim_(n→+∞) e^n^2 U_n

$$\mathrm{let}\:\mathrm{U}_{\mathrm{n}} =\int_{−\infty} ^{+\infty} \:\frac{\mathrm{cos}\left(\mathrm{nx}\right)}{\left(\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} }\mathrm{dx}\:\mathrm{calculate}\:\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \mathrm{e}^{\mathrm{n}^{\mathrm{2}} } \mathrm{U}_{\mathrm{n}} \\ $$

Question Number 135737 Answers: 2 Comments: 0

∫_(−2π) ^(4π) (3/(5−4cosx))dx

$$\int_{−\mathrm{2}\pi} ^{\mathrm{4}\pi} \frac{\mathrm{3}}{\mathrm{5}−\mathrm{4cosx}}\mathrm{dx} \\ $$

Question Number 135697 Answers: 0 Comments: 0

Question Number 135693 Answers: 1 Comments: 0

Ω = ∫ ((x−1)/((x−2)(x^2 −2x+2)^2 )) dx

$$\Omega\:=\:\int\:\frac{{x}−\mathrm{1}}{\left({x}−\mathrm{2}\right)\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}\right)^{\mathrm{2}} }\:{dx}\: \\ $$

Question Number 135673 Answers: 2 Comments: 0

∫ ((x^2 +1)/(x^4 +x^2 +1)) dx

$$\int\:\frac{{x}^{\mathrm{2}} +\mathrm{1}}{{x}^{\mathrm{4}} +{x}^{\mathrm{2}} +\mathrm{1}}\:{dx}\: \\ $$

Question Number 135646 Answers: 1 Comments: 0

∫(x+(1/2))ln(1+(1/x))−x dx=...?

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

Question Number 135633 Answers: 1 Comments: 0

∫(1/(x^(1/3) +1))dx=...?

$$\int\frac{\mathrm{1}}{{x}^{\frac{\mathrm{1}}{\mathrm{3}}} +\mathrm{1}}{dx}=...? \\ $$

Question Number 135662 Answers: 0 Comments: 0

Question Number 135627 Answers: 1 Comments: 0

... nice ................. calculus ... evaluation::::: 𝛗=^(???) ∫_0 ^( (π/2)) sin(x)ln(sin(x))dx solution::::: 𝛗=^(⟨cos(x)=y⟩) (1/2)∫_0 ^( 1) ln(1−y^2 )dy =−(1/2)∫_0 ^( 1) Σ_(n=1 ) ^∞ (y^(2n) /n)=((−1)/2)Σ_(n=1) ^∞ ((1/n)∫_0 ^( 1) y^(2n) dy) =((−1)/2)Σ_(n=1) ^∞ (1/(n(2n+1)))=−Σ_(n=1) ^∞ (1/(2n)) −(1/(2n+1)) =−((1/2)−(1/3)+(1/4)−(1/5)+...) =−1+(1−(1/2)+(1/3)−...) =−1+Σ_(n=1) ^∞ (((−1)^(n−1) )/n)=_(harmonic seties) ^(alternating) −1+ln(2) ∴ 𝛗= −1+ln(2)=ln((2/e))

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:...\:{nice}\:.................\:{calculus}\:... \\ $$$$\:\:\:\:\:\:\:{evaluation}:::::\:\:\:\boldsymbol{\phi}\overset{???} {=}\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} {sin}\left({x}\right){ln}\left({sin}\left({x}\right)\right){dx} \\ $$$$\:\:\:\:\:\:\:{solution}::::: \\ $$$$\:\:\:\:\:\:\boldsymbol{\phi}\overset{\langle{cos}\left({x}\right)={y}\rangle} {=}\:\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\mathrm{1}} {ln}\left(\mathrm{1}−{y}^{\mathrm{2}} \right){dy} \\ $$$$\:\:\:\:\:\:\:\:\:\:=−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \underset{{n}=\mathrm{1}\:\:} {\overset{\infty} {\sum}}\frac{{y}^{\mathrm{2}{n}} }{{n}}=\frac{−\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{n}}\int_{\mathrm{0}} ^{\:\mathrm{1}} {y}^{\mathrm{2}{n}} {dy}\right) \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{−\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}\left(\mathrm{2}{n}+\mathrm{1}\right)}=−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}{n}}\:−\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:=−\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{5}}+...\right) \\ $$$$\:\:\:\:\:\:\:\:\:=−\mathrm{1}+\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}−...\right) \\ $$$$\:\:\:\:\:\:\:\:\:=−\mathrm{1}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}\underset{{harmonic}\:{seties}} {\overset{{alternating}} {=}}−\mathrm{1}+{ln}\left(\mathrm{2}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\therefore\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\phi}=\:−\mathrm{1}+{ln}\left(\mathrm{2}\right)={ln}\left(\frac{\mathrm{2}}{{e}}\right) \\ $$$$\:\:\: \\ $$$$ \\ $$$$\:\:\:\:\: \\ $$

Question Number 135614 Answers: 0 Comments: 0

Question Number 135610 Answers: 1 Comments: 0

.... Advanced ...... Calculus.... prove that : determinant (((i :: Π_(n=0) ^∞ (1−(x^2 /((2n+1)^2 ))) =cos(((πx)/2)) ✓ )),((ii :: Π_(n=0) ^∞ (1+(x^2 /((2n+1)^2 )))= cosh(((πx)/2)) ✓✓))) .............

$$\:\:\:\:\:\:\:\:\:\:....\:\mathscr{A}{dvanced}\:\:......\:\:\mathscr{C}{alculus}.... \\ $$$$\:\:\:\:\:\:\:\:\:{prove}\:{that}\:: \\ $$$$\:\:\:\begin{array}{|c|c|}{{i}\:::\:\:\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\prod}}\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\right)\:={cos}\left(\frac{\pi{x}}{\mathrm{2}}\right)\:\:\:\:\checkmark\:\:}\\{{ii}\:::\:\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\prod}}\left(\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\right)=\:{cosh}\left(\frac{\pi{x}}{\mathrm{2}}\right)\:\checkmark\checkmark}\\\hline\end{array}\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:............. \\ $$

Question Number 135559 Answers: 2 Comments: 1

Question Number 135525 Answers: 0 Comments: 2

.... nice ................ calculus... evaluation of :: 𝛗=∫_0 ^( ∞) xe^(−x) (√(1−e^(−x) )) dx solution:: 1−e^(−x) =t ⇒ {_( x=−ln(1−t)) ^( e^(−x) dx=dt) 𝛗=−∫_0 ^( 1) ln(1−t).t^(1/2) dt =∫_0 ^( 1) Σ_(n=1) ^∞ (t^(n+(1/2)) /n)dt=Σ_(n=1) ^∞ (1/(n(n+(3/2)))) .... ∴ 𝛗= Σ_(n=1) ^∞ (1/(n(n+(3/2))))=Σ_(n=1) ^∞ (1/n)−(1/(n+(3/2))) .... =(2/3){γ−γ+Σ_(n=1) ^∞ ((1/n)−(1/(n+(3/2))))} .... we know that : ψ(s+1) := −γ+Σ_(n=1) ^∞ ((1/n)−(1/(n+s))) ..... ∴ 𝛗=(2/3)(γ+ψ((5/2))) .... on the oyher hand we have:: ψ(s+1)=(1/s)+ψ(s) ...... ψ((1/2))=−γ−2ln(2) ...... ψ((5/2))=(2/3)+ψ((3/2))=(2/3)+(2+ψ((1/2))) =(2/3)+2+(−γ−2ln(2)) .... ∴ ψ((3/2))=(8/3)−γ−ln(4) .... :::::: 𝛗 =(2/3)(γ+(8/3)−γ−ln(4)) .... 𝛗=(2/3)((8/3)−ln(4))=(4/3)((4/3)−ln(2)) .... ...... 𝛗= (4/3)((4/3)−ln(2))......✓✓ .... ... prepare by mr rizzy−aka... solution with detais :m.n.july.1970

Question Number 135513 Answers: 1 Comments: 0

Question Number 135495 Answers: 1 Comments: 0

hi, guyz ! let′s try this : I=∫_0 ^( 1) ((sin^2 x)/(cos^3 x))dx.

$$\boldsymbol{\mathrm{hi}},\:\boldsymbol{\mathrm{guyz}}\:! \\ $$$$\boldsymbol{\mathrm{let}}'\boldsymbol{\mathrm{s}}\:\boldsymbol{\mathrm{try}}\:\boldsymbol{\mathrm{this}}\::\:\boldsymbol{\mathrm{I}}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\boldsymbol{{sin}}^{\mathrm{2}} \boldsymbol{{x}}}{\boldsymbol{{cos}}^{\mathrm{3}} \boldsymbol{{x}}}\boldsymbol{{dx}}. \\ $$

Question Number 135443 Answers: 0 Comments: 0

Question Number 135389 Answers: 0 Comments: 0

let U_n =∫_(−∞) ^∞ ((cos(nx))/((x^2 −x+1)^2 ))dx calculate lim_(n→∞) e^n^2 U_n

$${let}\:{U}_{{n}} =\int_{−\infty} ^{\infty} \:\:\frac{{cos}\left({nx}\right)}{\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$$${calculate}\:{lim}_{{n}\rightarrow\infty} {e}^{{n}^{\mathrm{2}} } {U}_{{n}} \\ $$

Question Number 135382 Answers: 1 Comments: 0

find ∫_0 ^1 x^n ln(1−x^4 )dx with n integr natural

$${find}\:\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}} {ln}\left(\mathrm{1}−{x}^{\mathrm{4}} \right){dx}\:{with}\:{n} \\ $$$${integr}\:{natural} \\ $$

Question Number 135372 Answers: 0 Comments: 0

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

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

Question Number 135368 Answers: 1 Comments: 0

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

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

Question Number 135361 Answers: 1 Comments: 0

f(x)=3x^2 +6x,[−1,5] find−the−average−value

$${f}\left({x}\right)=\mathrm{3}{x}^{\mathrm{2}} +\mathrm{6}{x},\left[−\mathrm{1},\mathrm{5}\right] \\ $$$${find}−{the}−{average}−{value} \\ $$$$ \\ $$

Question Number 135259 Answers: 2 Comments: 0

∫(x^2 +3x)cos (x)dx

$$\int\left({x}^{\mathrm{2}} +\mathrm{3}{x}\right)\mathrm{cos}\:\left({x}\right){dx} \\ $$

Question Number 135257 Answers: 2 Comments: 1

∫t^7 ln(t)dt

$$\int{t}^{\mathrm{7}} {ln}\left({t}\right){dt} \\ $$

Question Number 135215 Answers: 0 Comments: 0

.....calculus preliminary.... Q: f(x)=2^x −2^(−x) ⇒ f^( −1) (x)=??? solution: y=2^x −2^(−x) ..... y=((2^(2x) −1)/2^x ) ⇒2^(2x) −y2^x −1=0 (∗) ... :: 2^x =t⇒ t>0 ...✓ .... (∗)→... t^2 −ty−1=0 .... Δ=y^2 +4>0...✓ ... t=((y+(√(y^2 +4)))/2) ...... :: 2^x =((y+(√(y^2 +4)))/2) ⇒_(both sides) ^(taking log) .... x:=log_2 (((y+(√(y^2 +4)))/2)) f^( −1) (x)=log_2 (((x+(√(x^2 +4)))/2)) ✓✓ .........................

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:.....{calculus}\:{preliminary}.... \\ $$$$\:\:\:{Q}:\:{f}\left({x}\right)=\mathrm{2}^{{x}} −\mathrm{2}^{−{x}} \:\Rightarrow\:{f}^{\:−\mathrm{1}} \left({x}\right)=??? \\ $$$$\:\:{solution}: \\ $$$$\:\:\:\:\:{y}=\mathrm{2}^{{x}} −\mathrm{2}^{−{x}} \:\:\:..... \\ $$$$\:\:\:\:\:\:{y}=\frac{\mathrm{2}^{\mathrm{2}{x}} −\mathrm{1}}{\mathrm{2}^{{x}} }\:\Rightarrow\mathrm{2}^{\mathrm{2}{x}} −{y}\mathrm{2}^{{x}} −\mathrm{1}=\mathrm{0}\:\:\left(\ast\right)\:... \\ $$$$\:\:\:::\:\:\mathrm{2}^{{x}} ={t}\Rightarrow\:{t}>\mathrm{0}\:...\checkmark\:.... \\ $$$$\:\:\:\:\:\:\:\left(\ast\right)\rightarrow...\:{t}^{\mathrm{2}} −{ty}−\mathrm{1}=\mathrm{0}\:.... \\ $$$$\:\:\:\:\:\:\:\:\Delta={y}^{\mathrm{2}} +\mathrm{4}>\mathrm{0}...\checkmark\:... \\ $$$$\:\:\:\:\:\:\:\:\:{t}=\frac{{y}+\sqrt{{y}^{\mathrm{2}} +\mathrm{4}}}{\mathrm{2}}\:\:\:...... \\ $$$$\:\:\:\:\:\:\:\:\:::\:\:\mathrm{2}^{{x}} =\frac{{y}+\sqrt{{y}^{\mathrm{2}} +\mathrm{4}}}{\mathrm{2}}\:\underset{{both}\:{sides}} {\overset{{taking}\:{log}} {\Rightarrow}}\:.... \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}:={log}_{\mathrm{2}} \left(\frac{{y}+\sqrt{{y}^{\mathrm{2}} +\mathrm{4}}}{\mathrm{2}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{f}^{\:−\mathrm{1}} \left({x}\right)={log}_{\mathrm{2}} \left(\frac{{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}}{\mathrm{2}}\right)\:\checkmark\checkmark \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:......................... \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\: \\ $$

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