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Question Number 166075 Answers: 0 Comments: 1

find the domain and range of the relation {(x,y):∣x∣+y≥2} by draw its graph

$$\mathrm{find}\:\mathrm{the}\:\mathrm{domain}\:\mathrm{and}\:\mathrm{range}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{relation}\:\left\{\left(\mathrm{x},\mathrm{y}\right):\mid\mathrm{x}\mid+\mathrm{y}\geq\mathrm{2}\right\}\:\mathrm{by}\:\mathrm{draw}\:\mathrm{its}\:\mathrm{graph} \\ $$

Question Number 166070 Answers: 0 Comments: 1

Question Number 166067 Answers: 0 Comments: 2

Question Number 166180 Answers: 0 Comments: 0

prove that 𝛗=∫_0 ^( 1) (( ln^( 2) (1−x ))/x^( 2) ) dx = 2 ζ (2) −−−proof−−− 𝛗= [((−1)/x) ln^( 2) (1−x) ]_0 ^1 −∫_0 ^( 1) ((2ln(1−x))/(x(1−x)))dx =−lim_( ξ→1^− ) (1/ξ)ln^( 2) (1−ξ)−2{ ∫_0 ^( 1) ((ln(1−x))/(1−x))dx+∫_0 ^( 1) ((ln(1−x))/x)dx} = −lim_( ξ→1^− ) {(1/ξ)ln^( 2) (1−ξ)+ln^( 2) (1 −ξ)}+2 ζ(2) =lim_(ξ→1^− ) (((ξ−1)/ξ))ln^( 2) (1−ξ) +2ζ(2) =_(ξ→1^− , δ→0^( +) ) ^(1−ξ= δ) [lim_( δ→0^( +) ) (((−δ)/(1−δ)))ln^2 (δ)=0] +2ζ(2) ■ m.n ∴ 𝛗 = 2 ζ(2)

$$ \\ $$$$\:\:\:\:\:\:{prove}\:\:{that} \\ $$$$\:\:\:\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:{ln}^{\:\mathrm{2}} \left(\mathrm{1}−{x}\:\right)}{{x}^{\:\mathrm{2}} }\:{dx}\:=\:\mathrm{2}\:\zeta\:\left(\mathrm{2}\right) \\ $$$$\:\:\:\:\:\:−−−{proof}−−− \\ $$$$\:\:\:\:\boldsymbol{\phi}=\:\left[\frac{−\mathrm{1}}{{x}}\:{ln}^{\:\mathrm{2}} \left(\mathrm{1}−{x}\right)\:\right]_{\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{2}{ln}\left(\mathrm{1}−{x}\right)}{{x}\left(\mathrm{1}−{x}\right)}{dx} \\ $$$$\:\:\:\:\:\:\:\:=−{lim}_{\:\xi\rightarrow\mathrm{1}^{−} } \frac{\mathrm{1}}{\xi}{ln}^{\:\mathrm{2}} \left(\mathrm{1}−\xi\right)−\mathrm{2}\left\{\:\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left(\mathrm{1}−{x}\right)}{\mathrm{1}−{x}}{dx}+\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left(\mathrm{1}−{x}\right)}{{x}}{dx}\right\} \\ $$$$\:\:\:\:\:\:\:=\:−{lim}_{\:\xi\rightarrow\mathrm{1}^{−} } \left\{\frac{\mathrm{1}}{\xi}{ln}^{\:\mathrm{2}} \left(\mathrm{1}−\xi\right)+{ln}^{\:\mathrm{2}} \left(\mathrm{1}\:−\xi\right)\right\}+\mathrm{2}\:\zeta\left(\mathrm{2}\right) \\ $$$$\:\:\:\:\:\:\:\:={lim}_{\xi\rightarrow\mathrm{1}^{−} } \left(\frac{\xi−\mathrm{1}}{\xi}\right){ln}^{\:\mathrm{2}} \left(\mathrm{1}−\xi\right)\:+\mathrm{2}\zeta\left(\mathrm{2}\right) \\ $$$$\:\:\underset{\xi\rightarrow\mathrm{1}^{−} \:,\:\delta\rightarrow\mathrm{0}^{\:+} } {\overset{\mathrm{1}−\xi=\:\delta} {=}}\left[{lim}_{\:\delta\rightarrow\mathrm{0}^{\:+} } \left(\frac{−\delta}{\mathrm{1}−\delta}\right){ln}^{\mathrm{2}} \left(\delta\right)=\mathrm{0}\right]\:+\mathrm{2}\zeta\left(\mathrm{2}\right)\:\:\:\:\blacksquare\:{m}.{n} \\ $$$$\:\:\:\:\:\:\:\therefore\:\:\:\boldsymbol{\phi}\:=\:\mathrm{2}\:\zeta\left(\mathrm{2}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\: \\ $$

Question Number 166063 Answers: 0 Comments: 2