53. (a)E‘0}x22x1dx1} 5limb→‘Eb0}x22x1dx1}5limb→‘3ln (x211)45limb→‘ln (b211)5‘Thus the integral diverges.(b)Both E‘0}x22x1dx1}and E02‘}x22x1dx1}must converge in orderfor E‘2‘}x22x1dx1}to converge.(c)limb→‘Eb2b}x22x1dx1} 5limb→‘3ln (x211)45limb→‘[ln (b211)2ln (b211)]5limb→‘050.Note that }x221x1}is an odd function so Eb2b}x22x1dx1}= 0.(d)Because the determination of convergence is not madeusing the method in part (c). In order for the integral toconverge, there must be finite areas in both directions(toward ‘and toward 2‘). In this case, there are infinite areas in both directions, but when one computes the integral over an interval [2b,b], there iscancellation which gives 0 as the result.54.By symmetry, find the perimeter of one side, say for 0 #x #1,y $0.y2/3512x2/3y5(12x2/3)3/2}ddyx} 5 }32}(12x2/3)1/212}23}x21/3252x21/3(12x2/3)1/21}d
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This note was uploaded on 10/05/2011 for the course MAC 1147 taught by Professor German during the Spring '08 term at University of Florida.