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I:=∫_0 ^( 1) ((ln (x))/(1 + x^( 2) )) dx := ∫_0 ^( 1) ln(x ) Σ_(n=0) ^∞ (−1)^( n) x^( 2n) dx := Σ_(n=0) ^∞ ( −1 )^( n) ∫_0 ^( 1) x^( 2n) ln( x )dx : = Σ_(n=0) ^∞ ( −1 )^( n) { [(x^( 2n+1) /(2n +1)) ln ( x )]_0 ^( 1) −(1/((2n +1 )^( 2) )) } : = Σ_(n=1) ^( ∞) ((( −1 )^( n−1) )/(( 2n +1)^( 2) )) = −G (Catalan constant )

$$ \\ $$$$\:\:\:\:\:\:\mathrm{I}:=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{ln}\:\left({x}\right)}{\mathrm{1}\:+\:{x}^{\:\mathrm{2}} }\:{dx} \\ $$$$\:\:\:\:\:\:\:\:\::=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \mathrm{ln}\left({x}\:\right)\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{\:{n}} \:{x}^{\:\mathrm{2}{n}} \:{dx} \\ $$$$\:\:\:\:\:\:\:\::=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\left(\:−\mathrm{1}\:\right)^{\:{n}} \:\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}^{\:\mathrm{2}{n}} \:\mathrm{ln}\left(\:{x}\:\right){dx} \\ $$$$\:\:\:\:\:\:\:\::\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\:−\mathrm{1}\:\right)^{\:{n}} \left\{\:\left[\frac{{x}^{\:\mathrm{2}{n}+\mathrm{1}} }{\mathrm{2}{n}\:+\mathrm{1}}\:\mathrm{ln}\:\left(\:{x}\:\right)\right]_{\mathrm{0}} ^{\:\mathrm{1}} −\frac{\mathrm{1}}{\left(\mathrm{2}{n}\:+\mathrm{1}\:\right)^{\:\mathrm{2}} \:}\:\right\} \\ $$$$\:\:\:\:\:\:\:\:\:\::\:=\:\underset{{n}=\mathrm{1}} {\overset{\:\infty} {\sum}}\frac{\left(\:−\mathrm{1}\:\right)^{\:{n}−\mathrm{1}} }{\left(\:\mathrm{2}{n}\:+\mathrm{1}\right)^{\:\mathrm{2}} }\:=\:−\mathrm{G}\:\:\left(\mathrm{Catalan}\:\mathrm{constant}\:\right) \\ $$

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Q :: # Calculus # If : 𝛗 ( n ) : = ∫_0 ^( 1) (( x^( 2n) )/(1 + x^( 2) )) dx then find the value of :: S := Σ_(n=1) ^∞ ((( −1 )^( n) 𝛗 ( n ))/n) = ? m.n.july.1970

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A region is enclosed by curves x^2 =4y, x^2 =−4y, x=4 & x=−4 V_1 is the volume of the solid obtained by rotating the above region round the y−axis. Another regions consists of points (x,y) satisfying x^2 +y^2 ≤16, x^2 +(y−2)^2 ≥4 and x^2 +(y+2)^2 ≥4 ,V_2 is the volume of the solid obtained by rotating this region round the y−axis Then V_1 =...

$$\mathrm{A}\:\mathrm{region}\:\mathrm{is}\:\mathrm{enclosed}\:\mathrm{by}\:\mathrm{curves} \\ $$$$\mathrm{x}^{\mathrm{2}} =\mathrm{4y},\:\mathrm{x}^{\mathrm{2}} =−\mathrm{4y},\:\mathrm{x}=\mathrm{4}\:\&\:\mathrm{x}=−\mathrm{4} \\ $$$$\mathrm{V}_{\mathrm{1}} \mathrm{is}\:\mathrm{the}\:\mathrm{volume}\:\mathrm{of}\:\mathrm{the}\:\mathrm{solid}\:\mathrm{obtained} \\ $$$$\mathrm{by}\:\mathrm{rotating}\:\mathrm{the}\:\mathrm{above}\:\mathrm{region}\:\mathrm{round} \\ $$$$\mathrm{the}\:\mathrm{y}−\mathrm{axis}.\:\:\mathrm{Another}\:\mathrm{regions} \\ $$$$\mathrm{consists}\:\mathrm{of}\:\mathrm{points}\:\left(\mathrm{x},\mathrm{y}\right)\:\mathrm{satisfying} \\ $$$$\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \leqslant\mathrm{16},\:\mathrm{x}^{\mathrm{2}} +\left(\mathrm{y}−\mathrm{2}\right)^{\mathrm{2}} \geqslant\mathrm{4}\:\mathrm{and} \\ $$$$\mathrm{x}^{\mathrm{2}} +\left(\mathrm{y}+\mathrm{2}\right)^{\mathrm{2}} \geqslant\mathrm{4}\:,\mathrm{V}_{\mathrm{2}} \:\mathrm{is}\:\mathrm{the}\:\mathrm{volume} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{solid}\:\mathrm{obtained}\:\mathrm{by}\:\mathrm{rotating} \\ $$$$\mathrm{this}\:\mathrm{region}\:\mathrm{round}\:\mathrm{the}\:\mathrm{y}−\mathrm{axis}\: \\ $$$$\mathrm{Then}\:\mathrm{V}_{\mathrm{1}} =... \\ $$$$\: \\ $$

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On souhaite calculer I=∫_0 ^∞ ((sint)/t)dt. (1) On de^ finit la fonction F(x)=∫_0 ^∞ e^(−tx) ((sint)/t)dt. (a) De^ terminer le domaine de de^ finition de f sur R. (b) Montrer que F est de classe C^1 sur R_+ ^∗ et calculer F ′(x). (c) Limite de F en +∞ ? Conse^ quence ? (2) On note Si(t)=∫_0 ^t ((sinu)/u)du pour tout re^ el t. (a) Montrer que G(x)=∫_0 ^∞ e^(−tx) Si(t)dt est de^ finie sur R_+ ^∗ . (b) Montrer que xG(x)→I quand x→0^+ . (c) Au moyen d′une inte^ gration par parties, montrer que F est continue en 0. (3) Calculer I.

$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{On}\:\mathrm{souhaite}\:\mathrm{calculer}\:\mathrm{I}=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}{t}}{{t}}{dt}. \\ $$$$\left(\mathrm{1}\right)\:\mathrm{On}\:\mathrm{d}\acute {\mathrm{e}finit}\:\mathrm{la}\:\mathrm{fonction}\:{F}\left({x}\right)=\int_{\mathrm{0}} ^{\infty} {e}^{−{tx}} \frac{\mathrm{sin}{t}}{{t}}{dt}. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{a}\right)\:\mathrm{D}\acute {\mathrm{e}terminer}\:\mathrm{le}\:\mathrm{domaine}\:\mathrm{de}\:\mathrm{d}\acute {\mathrm{e}finition}\:\mathrm{de}\:{f}\:\mathrm{sur}\:\mathbb{R}. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{b}\right)\:\mathrm{Montrer}\:\mathrm{que}\:{F}\:\mathrm{est}\:\mathrm{de}\:\mathrm{classe}\:{C}^{\mathrm{1}} \:\mathrm{sur}\:{R}_{+} ^{\ast} \:\mathrm{et}\:\mathrm{calculer}\:{F}\:'\left({x}\right). \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{c}\right)\:\mathrm{Limite}\:\mathrm{de}\:{F}\:\mathrm{en}\:+\infty\:?\:\mathrm{Cons}\acute {\mathrm{e}quence}\:? \\ $$$$\left(\mathrm{2}\right)\:\mathrm{On}\:\mathrm{note}\:{Si}\left({t}\right)=\int_{\mathrm{0}} ^{{t}} \frac{\mathrm{sin}{u}}{{u}}{du}\:\mathrm{pour}\:\mathrm{tout}\:\mathrm{r}\acute {\mathrm{e}el}\:{t}. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{a}\right)\:\mathrm{Montrer}\:\mathrm{que}\:{G}\left({x}\right)=\int_{\mathrm{0}} ^{\infty} {e}^{−{tx}} {Si}\left({t}\right){dt}\:\mathrm{est}\:\mathrm{d}\acute {\mathrm{e}finie}\:\mathrm{sur}\:{R}_{+} ^{\ast} . \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{b}\right)\:\mathrm{Montrer}\:\mathrm{que}\:{xG}\left({x}\right)\rightarrow{I}\:\mathrm{quand}\:{x}\rightarrow\mathrm{0}^{+} . \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{c}\right)\:\mathrm{Au}\:\mathrm{moyen}\:\mathrm{d}'\mathrm{une}\:\mathrm{int}\acute {\mathrm{e}gration}\:\mathrm{par}\:\mathrm{parties},\:\mathrm{montrer}\:\mathrm{que}\:{F}\:\mathrm{est}\:\mathrm{continue}\:\mathrm{en}\:\mathrm{0}. \\ $$$$\left(\mathrm{3}\right)\:\mathrm{Calculer}\:{I}. \\ $$

Question Number 144721 Answers: 0 Comments: 0

......... Nice ......∗∗∗......Calculus......... f ( x ) : = [ tan (x) + cot (x) ] R_( f ) = ? Hint:: [ x ] := Max { m ∈Z ∣ m ≤ x }

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:.........\:\mathrm{Nice}\:......\ast\ast\ast......\mathrm{Calculus}......... \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\mathrm{f}\:\left(\:\mathrm{x}\:\right)\::\:=\:\left[\:\mathrm{tan}\:\left(\mathrm{x}\right)\:+\:\mathrm{cot}\:\left(\mathrm{x}\right)\:\right] \\ $$$$\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{R}_{\:\mathrm{f}\:\:} \:=\:? \\ $$$$\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{Hint}::\:\:\:\left[\:\mathrm{x}\:\right]\::=\:\mathrm{Max}\:\left\{\:\mathrm{m}\:\in\mathbb{Z}\:\mid\:\mathrm{m}\:\leqslant\:\mathrm{x}\:\right\}\: \\ $$

Question Number 144720 Answers: 1 Comments: 0

.....calculus..... Ω := ∫_0 ^( ∞) ((sech(πx))/(1+4x^( 2) )) dx =^? (1/2) Ln(2)

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:.....\mathrm{calculus}..... \\ $$$$\: \\ $$$$\Omega\::=\:\int_{\mathrm{0}} ^{\:\infty} \frac{{sech}\left(\pi{x}\right)}{\mathrm{1}+\mathrm{4}{x}^{\:\mathrm{2}} }\:{dx}\:\overset{?} {=}\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{Ln}\left(\mathrm{2}\right) \\ $$$$ \\ $$$$ \\ $$

Question Number 144705 Answers: 1 Comments: 0

∫ (x^(n−1) /(x^(3n+1) (x^n −a))) dx ?

$$\:\:\int\:\frac{\mathrm{x}^{\mathrm{n}−\mathrm{1}} }{\mathrm{x}^{\mathrm{3n}+\mathrm{1}} \:\left(\mathrm{x}^{\mathrm{n}} −\mathrm{a}\right)}\:\mathrm{dx}\:? \\ $$

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........... Calculus........... In AB^Δ C : B^ = 2 C^ , a = λ b then specify the limits of the changes ′ λ ′ :

$$\:\:\:\:\:\:\:...........\:\:\mathrm{Calculus}........... \\ $$$$\:\mathrm{In}\:\:\mathrm{A}\overset{\Delta} {\mathrm{B}C}\:\:: \\ $$$$\hat {\mathrm{B}}\:=\:\mathrm{2}\:\hat {\mathrm{C}}\:\:\:\:,\:\:{a}\:\:=\:\lambda\:{b}\:\:\:{then}\:{specify} \\ $$$$\:{the}\:\:{limits}\:{of}\:{the}\:{changes}\:\:\:'\:\:\lambda\:\:'\:\:: \\ $$$$\:\:\:\:\:\:\:\: \\ $$$$ \\ $$

Question Number 144638 Answers: 1 Comments: 0

Triangle AOC inscribed in the region cut from the parabola y=x^2 by the line y=a^2 .Find the limit of ratio of the area of the triangle to the area of the parabolic region as a approaches zero

$$\mathrm{Triangle}\:\mathrm{AOC}\:\mathrm{inscribed} \\ $$$$\mathrm{in}\:\mathrm{the}\:\mathrm{region}\:\mathrm{cut}\:\mathrm{from} \\ $$$$\mathrm{the}\:\mathrm{parabola}\:\mathrm{y}=\mathrm{x}^{\mathrm{2}} \:\mathrm{by}\:\mathrm{the} \\ $$$$\mathrm{line}\:\mathrm{y}=\mathrm{a}^{\mathrm{2}} \:.\mathrm{Find}\:\mathrm{the}\:\mathrm{limit} \\ $$$$\mathrm{of}\:\mathrm{ratio}\:\mathrm{of}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{triangle}\:\mathrm{to}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{parabolic}\:\mathrm{region}\:\mathrm{as}\:\mathrm{a}\:\mathrm{approaches} \\ $$$$\mathrm{zero}\: \\ $$

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