From zeilberg@math.rutgers.edu Thu Oct 19 16:07:19 2006 Received: (from zeilberg@localhost) by math.rutgers.edu (8.11.7p2+Sun/8.8.8) id k9JK6Zi08013; Thu, 19 Oct 2006 16:06:35 -0400 (EDT) Date: Thu, 19 Oct 2006 16:06:35 -0400 (EDT) From: Doron Zeilberger Message-Id: <200610192006.k9JK6Zi08013@math.rutgers.edu> To: zeilberg@math.rutgers.edu Cc: lpudwell@math.rutgers.edu, scottsch@math.rutgers.edu Subject: MathIsFun; Today's "quiz" Status: RO Content-Length: 1092 Dear Students, Most people were on the right track, but only about a quarter got the right answer 4( sqrt(2) -1). Note that it is a number. Remember: sin(0)=0 sin(PI/2)=1 sin(Pi/4)=sqrt(2)/2 . Please review your elementary trig function evaluations! -----Solution to the "quiz"------- 1. Compute IntegralSign_{0}^{Pi/2} IntegralSign_{0}^{Pi/2} sin(x/2+y/2) dx dy Sol.: Recall the formulas Int sin(ax+b)= -cos(ax+b)/a +C Int cos(ax+b)= sin(ax+b)/a +C First Step: The inner integral is IntegralSign_{0}^{Pi/2} sin(x/2+y/2) = -cos(x/2+y/2)/(1/2) |_{0}^{Pi/2} = -2cos(x/2+y/2)|_{0}^{Pi/2}= -2[cos(Pi/4+y/2)-cos(y/2)]= 2cos(y/2)-2cos(Pi/4+y/2) Now do the outer integral IntegralSign_{0}^{Pi/2} IntegralSign_{0}^{Pi/2} sin(x/2+y/2) dx dy =IntegralSign_{0}^{Pi/2} [ IntegralSign_{0}^{Pi/2} sin(x/2+y/2) dx ] dy =IntegralSign_{0}^{Pi/2} [ 2cos(y/2)-2cos(Pi/4+y/2) ] dy= =2sin(y/2)/(1/2)-2sin(Pi/4+y/2)/(1/2) |_{0}^{Pi/2} =4[sin(y/2)-sin(Pi/4+y/2)] |_{0}^{Pi/2} =4 ( [sin(Pi/4)-sin(Pi/2)] - [sin(0) - sin(Pi/4)] )= 4(2sin(Pi/4)-1)= 4(sqrt(2)-1) . Ans: 4( sqrt(2) -1)