MAC 2312 Exam #la

Name: ~~~==--=- _~ K~-+--____

ID# _~~ Jo ~_-=---==--\ \ ___

HONOR CODE: On my honor, I have neither given nor received any aid on this examination.

Signature: ___________ ______

Instructions: Do all scratch work on the test itself. Make sure your final answers are clearly labelled. Be sure to simplify all answers whenever possible. SHOW ALL WORK ON THIS EXAM IN ORDER TO RECEIVE FULL CREDIT!!!

No. Score 1 /6 2 /10 3 /6 4 /6 5 /10 6 /6 7 /6 8 /18 9 /6 10 /26

Bonus /10 I Total I / 100 I

(1) Evaluate the integral. (6 paints)

Jy2 sin (6y) dy

(a) h 2 cos (6y) - Aysin (6y) - 168 cos (6y) + C

(b) -iy2 sin (6y) + /sycos (6y) + 168 sin (6y) + C

(c) - i y2 cos (6y) + ~y sin (6y) + ~ cos (6y) + C

@D-iy2cos (6y) + 118y sin (6y) + 168 cos (6y) + C¥ None of the above

S,\""(~~J; \A,1. -., ~-=-~ fA;

tA'V' :: <t.\'r\C,) 1 41\f ~ - i C$b(~)

~,t~·(,), ~ -i , 'c.os(,) 1- ~~ '1 cos(r,) '7 \;.( ~ck~~

rAtV -- {If. (<;), ~ IV -= 1; <btvd ~7)

~ l' tl" O.1~" :: - { yZCll\ (I.~l ;- {Lir ~l~(,) - USlh (1' ~J r~ 1.~1" (I;) ~ :: - i y' ClJ'> cr..,) +- ~7Si\.(~) - ~ (- ~ lOS (~l) t C

~'(SI,Jl.r' :: - -t{cos(,) -\- ti&"(~7) 1- ~ c.os(ryl1-C

(2) Evaluate the integral. (10 points)

xJe­ cos (4x) dx

u=- .e -)c ~ ettA ':: _e~>t k

ol/v-= t.o~(\h:) tAk ~ I\r =- -4 S1't\ ('\~)

Se-" Cll& ('(X) a.x '" ~ e-\"" ('t~) + ~ Je-~ S,,, ('1><)ol?c

-)c j -l(U~ e ~ vtl.1.-::-e ~

dAr-=- s,,, (I.h.)~ ~ I\r::- - ~ CDs(~)c)

I - 'Ie: , [ J. -)t () 1 r- '-1'If e <o\~ ('1 x).J-""q - '1 '( CoS "tlc - -q)e )Ccm (~~)O"X

it" (A', (4.,,) rh; ': ~ e-)t ~\~ ('1~) -ie -,\' cos ('-tx) - -t l~ -x c..os ~ It) oV.x

+r..kl (JI5 (4.~')o\Jx t *Sl-~ (.b!, ~~) oW

~Je-~ (j)~ (It,,) J« = ~ e-)C Si" (4..",) ~t -)< ~ ('\~)

*tv!)-\!'. To Sh k, -tt-;s "P'l'1o{l<o\ b k... 'i ...,hi) IA:: ~ ['Ix) t. d.,,-:o ( -"A, ~ ~~~tb ~~~ .

(3) Evaluate the integral. (6 points)

1~ cos2 (2x)sin3(2x) dx

2(a) io (b) is (c) 1~ ®15

'It,. jIt'/z.

~ lffi,' b.~) <&tn' ('b.) b. = • Cos' (2Jc) -=~ (ll<) Sill Cz><) ckJ \_ ~e. (2.x)

(4) Evaluate the integral. (6 points)

x212 v4 ­ dx

(a) 27f (c) ~ (d) ~

'SW\ e= J"t-;"L ~ ~ 4- '(.~ =: ~ ~he

Ct'X> e- ~ ~ ~ x.~ ;). toS 9

==:>ctIt == - ;) b\~{} de

= - Jet- J.( ~ s·", ttlf)) ~ - Jat-~ (;te-) a..sr" (f cns9

(5) Integrate the function, (10 points)

Jy'y2 - 25 ~- dy) y > 5

y3

-J-.. (1 0 ~~f{u- ~ 5"-'1l ~ e= "",,-'(Jy~r.) 5 5s CO~ &-: 1 ~ ~= c~(1 ~1=55eLe~(j='5tl.-'{~)~ COS~I(~)

-=;'i~= I'S~t~e

~o..,-= 5St~e-~&ole

rfyt-tr ~ ~ f 5~e _ (S5eC&-k.n&d&) :: J.r r~lG j&J r"J . I J\'L'i" ~eL~ e 5l S~c. l. e

=To e - -ra( ~ $,n(lf1))::. i~ I:) - ~~ ~t&~ ::. To 9 - i~ sw.G(j!;B B., ''It' e~G

----

(6) Integrate the function. (6 points)

J4X4 + 16x2 + 4 3 dx x +4x

(a) 2x2+In/x/-xIn/x2+4/+C (b) 2x2-~In/x2+41+C

~ 2x2+ In lx/ - -2

1 In Ix2+ 4/ + C () 2~ d 2x + In /xl + ~ In Ix2+ 41 + C (e) None of the above

\f

,-~'1f\\' ) alA: = IfJX tk + L(~,~) &k ~

2'(..2

A (~l..+'1) 4- ~ (BtfC.)

X(~;t~) X(lI:2.~)

~(\Ltli)+ X (B~+()=: '1

)( ~o : 4A -::. \.( ~A-::::-I

21\~+"f + 13y;2 +tlt -::. 'i ~ ('e> i-i) X +-(~ +I.f :=. ~

~+- 1=0 ~ 8::.-1

c. -=-0

~ l<Cx~ftM~ f( ~ -~)~ ~ ffc4: -S/¥t ~ u -::. )1;.2.+'1

~ ':: ~kM -7 +c&-= ~04-

= 111\ I~ \' i: r ~ d.M = \~ l)t /- i I~ It.\ ' r '\X Y -\--I~~a.+,-\

2 -=- I" I~1- ~hI,,~~ J

j X~ +- "t )c ~ =: 20\ + l~ \~ \ - ~ I", IX'l fl41 +(

(7) Integrate the function. (6 points)

Jx2 (x; _ 9) dx

(a) 9~ + 5~ In Ix + 31 - 514 In Ix - 3/ + C

(b) 9~ + AIn Ix - 31- 2~ In Ix + 3/ + C

J& l~X + .5~ In Ix - 31 - 5~ In Ix + 31 + C

~9~ + i4 In Ix - 31 - 5~ In Ix + 31 + C (e) None of the above

___ - , JA.. B C \2.t~~-") \~(~~~'Cx-~)::)Z'+ .".'2. ~ + \~3-4­

_ A)C ('l.:tl)(~ -l) ~ 'B ()a; +1)();-'l) + C)c~ (~ -3) +1>..\1. (}t;+3)

'(.,'2. (~+3) ('J,-~)

\ -::: A- X (~+3)(~-3) +-'\3> eXt1)C~-3) + C~1 (l<: -3) + l>~? (~+J)

x~o : \-:;:-(H)=9B~- ~ x: = -3~ \-=-5l{( =>c:::- ~

x:::.~ : \::. S\-(t ~ 1) ~ ~

I~ ~)c (lc~~) _ *C~2~~) _ ~ (\3_~2-)+ i (~~ -+3~2.)

lo:: A-.3 -1A1. - -V+-l- ~ of- ¥~ ¥ -<- W l ~fu~ -1kx tl ~ ,4--=-0

-;.4 =-0

Lt[)J,j eM =- ~ ~ t o\oc - sol- h:} ~ + ~-h~l '~

-::: - t t~ \~l- ~ l~ IX+3, +- ~ lr\ Ik-3) +C

(8) (i) Use the Trapezoid Rule with n = 4 steps to estimate the integraL (6 points)

126x2 dx e3 (b) 16 (c) 33 (d) ~ {

A"' :: ~ :: '2. - 0 - 2.. _ .1 A ~ -q-- - ,, - 2

)c.c -= 0 ::::!) .f (f»): ~(O),. =("(1>\:'0

't I -:: 0 4 -i :: ~ ~ + ( i ) ~ (" (t')~ :0 '" ( ~) ~ ~

~"Z:: ~+~ ~ I :::, +' (t) ':" ~ ( \ )"2 -= C. ( I ) ~ '"

'(3 ~ ).. i ~ { "9 -r(-l)~ b(!y ~ ~(~) ~2-;

'('1 -=: 2 ~ +(2.\= <a C~)'Z- == C. ('1) ~2'i

r b",' oI.A: .,. 6'; [-+(~;)~ 2 -I (~,) n .f(~.\! 2-r(k~ +- -t6·~)J o ::. ..l [b-l-.;l.(-l) +;(!.)-I-JC"j+ 2't1 ~ ~[e~ ~IIU nH~ > ~[(h1

~ - n 2

(ii) Find an upper estimate for the error in estimating

1: (2X 3 + 9x) dx

using the trapezoid rule with n = 6 steps. (6 points)

(a) ~4 @D 1~5 (c) 2~O (d) 1;;

ftk.)-': 2'1.." ~

f\(~}-= Co",,+~

{\\ (~):. \1)(

~l\r\ ('()\1: \1(~)~~\

(iii) Estimate the minimum number of subintervals needed to approximate the integral

3 21 (3x + 1) dx

with an error of magnitude less than 0.0001 using the Trapezoid Rule. (6 points)

(a) 71 (b) 347 (c) 100 @ 200

+{~\~ ~t~+\

.f l('(.)::: (,.X

{\tX.):: l6

~ l~"l~)\ == lo

\e...\ L _~~l) ')

l~) f C) ,Cool \ - 1'2-\'\'

:, d l~) ~ o,o~\ \'2",1

~ 0.000\

'in. L 0.000\ I -jx.'l- - .

l ~O,ooOl",2.­

- "-\ ~V)La,()OO\

Y()OOo ~",z..

1J)o '=..-V'

(9) Evaluate the improper integral or state that it is divergent. (6 points)

100

8e-8x dx

(a) Divergent @ 1 (c) -1 (d) 0 (e) None of the above

(10) (i) Evaluate the improper integral or state that it is divergent. (9 points)

11 1 - dx

-1 x~

(ii) Determine whether the improper integral is convergent or divergent. Justify your answer. (8 points)

1= 2 + sinx 2 dx

7f X

I;

(iii) Evaluate the improper integral or state that it is divergent. (9 points)

1 200

--dx X 22 - 1

2­- __I ­

(k+'M~-\)

AtX..... l) +0 (lttl) ==- Z

'(::.\ '. 2e = 2~B~'

x:=..- \ : -2A-=Z. ~A-:: - '

J x& oW " ) (- ± + ~I_,) Jx = - \" \x+d + , .. Ix-I \ 00 \" 1:j

- ~:. ~ \~ ~\ - \~ \~ \1 ~ - \W\ \ ~ \ ~ [0 ~

~\.)

¥ ~()-k:To ~ ~~ ~~ F~\tWV1 ',~ cLw ~ -tn, Of\()~'i( sV'o!,-h-hJ-\)~ 1

S& £){Am lb ~\VV ~~.

- -

I

Bonus. Evaluate the improper integral or state that it is divergent. (10 points)

100

e- ..fi dx

~-=- (X

dw. -=- -1 )<.... 1 \

M ~ 2.[i ctu.=rut ~ 2lA dM. ~ rJvx. '-""""' ~

) t - (ii c4. :: 7. ~ \A e-" ctv..

'f ~ \.\ ~ dvt ::.tk., - \..\

d.w -::. t-'\J». -4 I\f ~ - -e

- i't ­e