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Question Number 165273 by mnjuly1970 last updated on 28/Jan/22
limn→∞∫01n.e1−x21+n2x2dx=?−−−−−−
Answered by mindispower last updated on 28/Jan/22
∫n1+n2x2dx=arctan(nx)limn→∞[arctan(nx)e1−x2]01+2∫01xe1−x2arctan(nx)dx=lim2n→∞∫01arctan(nx).xe1−x2dx2∫01xtan−1(nx)e1−x2dx⩽2∫01π2xe1−x2=π2weUse‘‘TheoremdeConergenceDominee″itsTheNameinfranceinEnglishIdontKnowHowWeCalliTlim2n→∞∫01xtan−1(nx)e1−x2dx=∫01lim2tann→∞−1(nx).xe1−x2dx=∫01π2(2xe1−x2)dx.π2[−e1−x2]01=π2e
Commented by mnjuly1970 last updated on 28/Jan/22
bravo,sirpower....verynicesolutionDominatingconvergencetheorem...
Commented by mindispower last updated on 02/Feb/22
ThanxSirHaveaNiceDay
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