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WUSTL MATH 132 - m132_E1sF10

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MATH 132 EXAM I FALL 2010(Solutions) 1) Estimate the area under the graph of = from x = to x = 0ÐBÑ % ! B ! ###using 4 approximating rectangles and midpoints (i.e. using M )%ÞEÑ "! FÑ "" GÑ "# HÑ "$ IÑ "% J Ñ "& KÑ "' LÑ "( MÑ ") N Ñ "*Solution: ?Bœ œ"ÞQ œ"Ð0Ð! "Þ&Ñ " 0Ð ! Þ&Ñ " 0ÐÞ&Ñ " 0 Ð"Þ&ÑÑ œ#!Ð!#Ñ%%"Ð"Þ(& " $Þ(& " $Þ(& " "Þ(&Ñ œ "" ÐFÑ--------------------------------------------------------------------------------------------2) Which of the following limits represents the area under , 2 4 ?0ÐBÑ œ B Ÿ B Ÿ10 2EÑ Ð# " ÑFÑ Ð#" ÑGÑ Ð" Ñlim lim lim8Ä_8Ä_8Ä_"3"""3" #3""88 88 8 83œ! 3œ" 3œ!8!"8 8!"!! !10 10 1022HÑ Ð " ÑIÑ Ð"" ÐÑÑJÑ Ð"" Ñlim lim lim8Ä_8Ä_8Ä_88 8 8 8 83œ" 3œ! 3œ!88!"8!"3"3""#3#22 210 10!! !2 KÑ Ð" " ÑLÑ ÐÑMÑ Ð#" Ñlim lim lim8Ä_8Ä_8Ä_#3"""3"""388 88 883œ! 3œ! 3œ!8!"8!"8!"!!!10 10 102NÑ Ð"" Ñlim8Ä_"3""883œ!8!"#!Solution: For The first term in the summation will be Vß Bœ œ Þ Ð#" ÑÞ8%!## #88 8"!? (D) is the one that satisfies both properties.---------------------------------------------------------------------------------------------3) represents which of the following lim8Ä_88 83œ8'3 '3#61!’“Ð# " Ñ " #Ð# " Ñ definite integrals ?EÐB" #BÑ .B FÑ ÐB " #Ñ .B GÑ Ð " #BÑ.B H Ñ Ð#B " BÑ .B)x''''!# %### #)$ "!6-3IÑ ÐB " #BÑ .B J Ñ Ð#B " %BÑ.B KÑ Ð#B " %BÑ.B LÑ Ð#B " %BÑ .B '' ' '## ! #)) ' )### #Solution: = 6 The first term in the summation?Bœß=9,! +œ Þ 3=,!+'88Ð# " Ñ " #Ð# " ÑÞJ9< +œ#ß ,œ)ß 0ÐBÑœB " #Bß ÐIÑ''88## and we get , satisfyingboth. None of the others work.#Þ4) where the term inside the square bracket is R ,'’“!+,8Ä_Ð,!+Ñ83œ"8380ÐBÑ.B œ 0ÐB Ñ ßlim the Riemann sum with being the right endpoint. Using the formulaB3 , we have R where R!'3œ"8##8Ð8""ÑÐ#8""Ñ'!$8Ä_883œ B" #.Bœ ß œlimEÑ FÑ GÑ HÑ#$ * #$8 #8 #8 *88Ð8""ÑÐ#8""Ñ 8Ð8""ÑÐ#8""Ñ 8Ð8""ÑÐ#8""Ñ 8Ð8""ÑÐ#8""Ñ$$$ $IÑ " JÑ " KÑ " '&(*# 8 #8 #88Ð8""ÑÐ#8""Ñ 8Ð8""ÑÐ#8""Ñ 8Ð8""ÑÐ#8""Ñ$$$10 8LÑ " $MÑ " &NÑ " (68 108Ð8""ÑÐ#8""Ñ 8Ð8""ÑÐ#8""Ñ 8Ð8""ÑÐ#8""Ñ88 8$$ $Solution: , and So Then ,!+ $ $3 $388 8 833 3#3œ"8œBœÞ0ÐBÑœÐÑ" #Þ 0ÐB Ñ œ!** ,!+8'8 83œ" 3œ" 3œ"88 8#8Ð8""ÑÐ#8""Ñ83##!! !3 " #œ " #8Þ œ 0ÐB Ñ œHence R $* $ *8' 8 8 # 88Ð8""ÑÐ#8""Ñ 8Ð8""ÑÐ#8""Ñ#$" #8 œ " 'Þ ÐKÑ----------------------------------------------------------------------------------------5) Evaluate the integral 2 by interpreting it in terms of areas'!6lB! %l.Bß ÞEÑ FÑ GÑ HÑ IÑ J Ñ KÑ LÑ MÑ N Ñ46 8 101214 16 182022Solution: = ''!##'Ð% ! #BÑ .B " Ð#B ! %Ñ .B œ % " "' œ #!Þ ÐMÑ----------------------------------------------------------------------------------- .6) Find a particular solution for the indefinite integral sec ( sec + tan ) .'!! !!.EÑ =/- FÑ =/- >+ 8 GÑ >+8 " =/- H Ñ =/- IÑ# #"#!!!!! ! !2 secJ=/- " >+8 KÑ >+ 8 LÑ >+8 =/- MÑ >+8 N Ñ # >+8)## # #""##!! ! !! ! !Solution: = ( sec + sec tan ) tan sec '#!!!! ! !.œ ""GÞ ÐGÑ---------------------------------------------------------------------------------------7) If then find 2ÐBÑ œ > " "' .>ß 2 Ð"ÑÞ'È#B "#B#w#EÑ #( FÑ #' GÑ #& HÑ #% IÑ #$ J Ñ ## KÑ #" LÑ #! MÑ "* N Ñ ") Solution: By Fund Theorem, Then2ÐBÑœ ÐB " #BÑ " "' Ð#B " #ÑÞw##È =2 Ð"Ñ #& Ð%Ñ œ #!Þ ÐLÑwÈ---------------------------------------------------------------------------------------8) Using the substitution 1 + tan(t) write the integral ?œ ß .>ß'!Ð>Ñ "" >+8Ð>Ñ1/41cos#Èas an integral in the variable u .A) '' ' '' ''È!! ! !! """" " Î%##.? .? " .? .? .? .?? ??#? ? ?È ÈFÑ # ? .? GÑ HÑ I Ñ J Ñ KÑ11##LÑ ? .? MÑ # ?.? NÑ'' 'ÈÈ"" "## #.??#Solution: , , and .? œ =/- Ð>Ñ .> œ .> > œ ! Ê ? œ " > œ Ê ? œ #Þ#"-9= Ð>Ñ %#1Then ''!"Ð>Ñ "" >+8Ð>Ñ#.??1/41cos#ÈÈ.> œ Þ ÐKÑ3.9) Evaluate the integral '0/41"!-9=Ð#>Ñ#.>ÞEÑ FÑ GÑ HÑ IÑ J Ñ " #KÑLÑMÑ#NÑ!111 1 1 1!# "" " %#!"#" $")#% ) %# %111Solution: ''00/4 /411"!-9=Ð#>Ñ =38Ð#>Ñ####%"" >.> œ ! -9=Ð#>Ñ .> œ ! lœ1Î%!Ð ! Ñœ Þ ÐEÑ11)% )" !#----------------------------------------------------------------------------------------------------10) Find the between the curve x and the x-axis, on [0,3].total area 0ÐBÑ œ ! #B# : The is area below the x-axis (area above the x-axis).Recall total area ÐÑ"EÑFÑGÑ"HÑIÑJÑ#KÑLÑMÑ$NÑ"# %& ( ) "!$$ $$ $ $ $Solution: The area below the x-axis is for 0 x 2. That area is ŸŸ ß! ÐB ! #BÑ .B'!##œÐ#B! BÑ.Bœ Þ #Ÿ Ÿ$'!##%$The area above the x-axis is for x . That area is'#$#%%%)$$$$B ! #B .B œ Þ " œÞÐLÑTherefore the total area = ------------------------------------------------------------------------------------------------11) Evaluate the integral to 4 decimal places'!"Î#! >"! >1È#.> ß ÞEÑ FÑ GÑ HÑ IÑ J Ñ K Ñ0.2576 0.3896 0.4765 0.5143 0.5896 0.6854 0.7453LÑ %$ MÑ NÑ0.79 0.8642 0.9468Solution: = The first is arcsin(t)| The second is''!!"Î# "Î#"! >"! >! >'1ÈÈ##"Î#!.> " .>Þ œ Þ1 by the substitution, Then ?œ"! >ß.?œ! #> .>Þ .> œ .? œ#!""Î# $Î%! >""! >#?''ÈÈ#È'?l œ ! "Þ .> "!" œ Þ$)*'#% µ Þ$)*'Þ ÐFÑ$Î%"#ÈÈÈ$$#'#!"Î#! >"! >Then = 1 1-----------------------------------------------------------------------------------------------------12) Evaluate the integral where , and b are positive constants.'!"+B+B " ,4( )##.Bß +A) D) "#%"" ##%%+",+",+",,+",,+",,+",FÑ GÑ ! IÑ ! JÑ !KÑ LÑ MÑ NÑ"# # %+++!,+!, Solution: Substitution Then =?œ+B " ,ß .? œ #+B.BÞ .B#!"+B+B " ,'4( )## ',+",####??,+",#+",,.? œ ! lœ! ÞÐIÑ-------------------------------------------------------------------------------------------------13) On what interval is the curve ?Cœ0ÐBÑœ .>ß'!B>" " >#concave upwardEÑ Ð!ß _Ñ FÑ Ð ! _ß !Ñ GÑ Ð!ß "Ñ HÑ Ð ! "ß !Ñ IÑ Ð ! "ß "Ñ J Ñ Ð!ß #ÑKÑ Ð ! #ß !Ñ LÑ Ð ! #ß #Ñ MÑ Ð ! #ß #Ñ N Ñ Ð ! _ß _ÑSolution: By Fund. Theorem, By quotient rule f0ÐBÑœ Þ ÐBÑœ ÞwwwB"!B""BÐ""BÑ#### f for so that is where f(x) is concave upward.


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WUSTL MATH 132 - m132_E1sF10

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