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Question Number 165273 by mnjuly1970 last updated on 28/Jan/22

        lim_( n→ ∞)  ∫_0 ^( 1) ((  n . e^( 1− x^( 2) ) )/( 1 + n^( 2)  x^( 2) )) dx =?            −−−−−−

limn01n.e1x21+n2x2dx=?

Answered by mindispower last updated on 28/Jan/22

∫(n/(1+n^2 x^2 ))dx=arctan(nx)  lim_(n→∞) [arctan(nx)e^(1−x^2 ) ]_0 ^1 +2∫_0 ^1 xe^(1−x^2 ) arctan(nx)dx  =lim_(n→∞) 2∫_0 ^1 arctan(nx).xe^(1−x^2 ) dx  2∫_0 ^1 xtan^(−1) (nx)e^(1−x^2 ) dx≤2∫_0 ^1 (π/2)xe^(1−x^2 ) =(π/2)  we Use “Theorem de Conergence Dominee”  its The Name in france in English I dont Know  How We Call iT  lim_(n→∞) 2∫_0 ^1 xtan^(−1) (nx)e^(1−x^2 ) dx=∫_0 ^1 lim_(n→∞) 2tan^(−1) (nx).xe^(1−x^2 ) dx  =∫_0 ^1 (π/2)(2xe^(1−x^2 ) )dx.(π/2)[−e^(1−x^2 ) ]_0 ^1 =(π/2)e

n1+n2x2dx=arctan(nx)limn[arctan(nx)e1x2]01+201xe1x2arctan(nx)dx=lim2n01arctan(nx).xe1x2dx201xtan1(nx)e1x2dx201π2xe1x2=π2weUseTheoremdeConergenceDomineeitsTheNameinfranceinEnglishIdontKnowHowWeCalliTlim2n01xtan1(nx)e1x2dx=01lim2tann1(nx).xe1x2dx=01π2(2xe1x2)dx.π2[e1x2]01=π2e

Commented by mnjuly1970 last updated on 28/Jan/22

  bravo , sir power ....very nice   solution      Dominating  convergence theorem...

bravo,sirpower....verynicesolutionDominatingconvergencetheorem...

Commented by mindispower last updated on 02/Feb/22

Thanx Sir  Have  a Nice Day

ThanxSirHaveaNiceDay

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