Test Review with solutions on improper integrals - ali(zra89 FINALReview gilbert(55035 This print-out should have 24 questions Multiple-choice questions

Test Review with solutions on improper integrals -...

This preview shows page 1 - 4 out of 14 pages.

ali (zra89) – FINALReview – gilbert – (55035)2Consequently, the given sequencedoes not converge.0030.0pointsLetgbe a continuous, positive, decreasingfunction on [5,).Compare the values ofthe integralA=integraldisplay145g(x)dxand the seriesB=14summationdisplayn=6g(n).1.A < B2.A=B3.A > BcorrectExplanation:In the figure56789. . .a6a7a8a9the bold line is the graph ofgon [5,) andthe areas of the rectangles the terms in theseriessummationdisplayn=6an,an=g(n).Clearly from this figure we see thata6=g(6)<integraldisplay65g(x)dx,a7=g(7)<integraldisplay76g(x)dx ,whilea8=g(8)<integraldisplay87g(x)dx,a9=g(9)<integraldisplay98g(x)dx ,and so on. Consequently,A > B.keywords: Szyszko0040.0pointsDetermine the interval of convergence ofthe seriessummationdisplayn=1n2n(x5)n.1.interval convergence = [2,5 ]2.interval convergence = [2,5 )3.interval convergence = [ 3,7 )4.interval convergence = [ 3,7 ]5.interval convergence = ( 3,7 )correct6.interval convergence = (2,5 )Explanation:Setan=n2n(x5)n.
ali (zra89) – FINALReview – gilbert – (55035)3Thenan+1an=2n(n+ 1)2n+1n(x5)n+1(x5)n=(n+ 1)(x5)2n=x52parenleftbigg1 +1nparenrightbigg.Butlimn→ ∞parenleftbigg1 +1nparenrightbigg= 1,solimn→ ∞vextendsinglevextendsinglevextendsinglevextendsinglean+1anvextendsinglevextendsinglevextendsinglevextendsingle=12|x5|.Thus the given series(i) converges for|x5|<2, and(ii) diverges for|x5|>2,and so converges on the interval (3,7) anddiverges outside [3,7].It remains, therefore, to check convergenceat the endpoints|x5|= 2,i.e., atx= 3,7.Now atx= 3, the series becomessummationdisplayn=1(1)nn ,while atx= 7 it becomes

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture