AllWorksheets125Aut2011

AllWorksheets125Aut2011 - 2 x 4 6 § † œ„ ´ ¥ † “„...

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Unformatted text preview: 2 x 4 6 § † œ„ ´ ¥ † “„ ´ xq f u v f u x ’ d o h ’q  u v u d t u ƒ f d ‡%–r¦Y™r¢s#wi¤w8C¢is8EwB‚F¦‚!˜˜ † ‰ „ fv ’† f sf‚žy8ˆn¡„ ‘ † …„ ’ˆr´ £‰ f v u h u  hq £ ƒ h !FC8s˜`‚if 2 C(x) ‘ †œ„ ¤–r´ ‘ † Ÿ„ 8 r´ ‘ † “„ ™r´ ‘ † ¢„ x¡r´ 4 h(x)=6-x 6 °†ƒ!F8isf‚B‚F¦E 3j!¦Css—!”swyE8¦‚F‚i¯svŽ• h f f v  q ƒ u d t u u ˜„ h … t f u hq‰ ‰ f ‘q t v u l u d t ƒ h f xq u f u d h u u ot u • … ¥ ¸ ‘ † …„ ¶ h ’q t ‘ h „† u d t ƒ h f q u f CFt‹E“pET®ºg¹rˆ·8swE‚CˆCF¦‚in€xC¤Ž‚… uCFŒpx•‹‚‚C8‹Š¤‚ƒ‚F™tFžˆrµ‚‚Eu  ”' dt ƒ u ƒ h „ ’ • ‡ ‰ ˆ ‡ u dt u • ’t † …„ ´ u h — D § † œ„ ¨ ¥ † “„ ¨ xq f u v f u x ’ d o h ’q  u v u d t u ƒ f d ‡%––zY¡–¢s#wi¤w8C¢is8EwB‚F¦‚!˜˜ † f‰ „ f v ’† f ©‚ew8ˆž™„ 1 x 2 3 4 † u‰  h f t ‘ u v ¬ u‰  h fq v t w hq d t ² t hq ± ”s8Ci&‡w³Bs8Ci©yFrCs‚w2”sq™„ ‘ † …„ k–¨ £‰ f v u h u  hq £ ƒ h !FC8s˜`‚if 1 ‘ †œ„ ¤Ž–¨ ‘ † Ÿ„ 8 …¨ ‘ † “„ ¡–¨ ‘ † ¢„ x¡…¨ 2 °†ƒ!F8isf‚B‚F¦E 3j!¦Css—!”swyE8¦‚F‚i¯svŽ• h f f v  q ƒ u d t u u ˜„ h … t f u hq‰ ‰ f ‘q t v u l u d t ƒ h f xq u f u d h u u ot u • … ¬ œ ‘ † …„ ª h q t ‘ h „† u d t ƒ h f q u f CFt®E¨p‡‹­¦«(`Œ8€’F‡‚‚ˆ–‚Fz‚ig€xC¤Ž‚… uCFzp•‹C‚‚8T‹x¤‚™‚F©Fžˆ…g‚‚Eu  i' dt ƒ u ƒ h „ ’ • ‡ ‰ ˆ ‡ u dt u • ’t † …„ ¨ u h —  § † œ„ ‚ ¥ † “„ ‚ xq f u v f u x ’ d o h ’q  u v u d t u ƒ f d ‡¤Ž#¦Y¡#js#wi¤w8C¢is8EwB‚F¦‚!˜˜ 3 B(x) g(x)=1+x † f‰ „ fv ’† f 8©‚žy8ˆ#™„ ‘ † …„ ’#‚ £‰ f v u h u  hq £ ƒ h !FC8s˜`‚if 1 2 x 3 4 ‘ †œ„ ¤Ž#‚ ‘ † Ÿ„ 8 ž‚ ‘ † “„ ¡#‚ ‘ † ¢„ x¡ž‚ A(x) ›†™iw8i©CšCF¦™3j!¦‚€hs—isF–iBCF‚if h f f v  fq ƒ u d t u u ˜„ h … t f u q‰ ‰ f ‘q tv u l u d t ƒ h sxvpxB‚wƒE¨p‡#”’2™8swE‚CˆBCF–‚i¢€xC¤Ž‚… q u f • u d t h u u ot u • “ ‘ † …„  h ’q t ‘ h „† u d t ƒ h f q u f u dt p • ƒ u ƒ h „ ’ • ‡ ‰ ˆ ‡ u dt u • ’t † …„ ‚ u h — CFŒx‹‚‚C8‹Š¤‚ƒ‚F™tF#ˆ#ƒ‚˜‡u  %'  € I a Q X I W ~ | {S YdcUb`Y¨¢}&zy h t u u d x w v ’ o xq d t hq q d x h ’q t f‰ u EE‚wxwi¨jsCF…s–€m‚wC8sw!©Ewv t f d t u v ‰ m u u q o u s h h q t ‘ h „† f u v f u d t ƒ h f h q t ‘ h „† ‰ f q q v ’ u d t h u u ot u • mq d x h q t f‰ u v u ‘q h p v u l f v˜Fjwis’‚‡˜€‰s‰¢jtYr8€’FEC‚ˆBwi3‚wr‚ir8€’FEC‚ˆi‚sh8€w8q‚Fr“pETns‚y‚8€’F!sw¢Es‚gyE8¦ksxq u v u d i h h q t ‘ h „† v u d t ’ u f ’ x v u ƒ h „ f u v f u d t x f ƒ u h — u ƒ u • h f ‘ x h ’q t ‘ h „† ‰ „† u x „ ƒ h f  q t x u v u t hq p h f y‚j2`8s’wE‚‚ˆ“‚widg8w¨‚‚Ce”yi6‚w“i™E‚˜ECdT–i”“‚8swE‚Cˆ‚ˆ‡w‚…Ci#‚€hFyEwEFsrige 3 f(x)=3 IdcUabQ`XYW¦V¢U&TRPH SQI £ 321 G 7$D£ A 75 @[email protected] 0)' ¡$ !%(&%#" !£ ©©¨¦¤¢   §   £§ ¥ £ ¡ É u t f fq tx uv u ƒ h %w!rsF–P–C‚‚„ h f v ’ u t f q t x uv u l ’ h f q xà h † Ÿ„ ‰ ƒ h f £ u lv „ ‘ u d t v u ƒ h „ f u v f u d t † ’ h q t f fq u ’ v m m f t x v — v „ ’ p ini…F!rsfwyEp–E8%zinctgpt‡8™shz‚!“!8–‚–‚w#‚‚C×wiCFqi½8€’F!rs”wC‚iryy˜ž‚8ižsxq x€‚jBr‚ß‚wµ‚iׂ!¦€u8‚iFEy…ž‚wi6iwinCFƒCshYšEs8Ci&‡we‚FidF8sE‚ž‚weF‚ri‘ q d i h m „ f u d t ƒ ƒ f ƒ h f x ‰  h f t ‘ u v ¢ u d t † ’ x f u v f u d t ƒ q » h x u‰  h f t ‘ u v u d t † ’ x t d q u d u d t u t „ m f ’ ’ t … Ó œ ‘ † …„  ’ t q x u t f q ƒ v ’ ’ ‘ … u x u d t  ‰ Þ É x ‰  h f t ‘ u v u d t † ’ x u ƒq x u d t † ’ x u t f q ƒ v ’ ’ ‘ … u d t u v w©¤¤ÊdˆY™Fsh6EF!‚sh‚y8Ep‚nwE‚F…‚€„¿Fsui‚i&‡y#‚Fki¦Csw#CFki6w!˜€h‚w8ŠP‚ž‚Fžyif t f d s h d m f v  v „ ’ p h ’ x ‰  h f t ‘ u v ¢ q d ‘ t u w x £ t d q u d u d t x f u ‘q‰ x d ‘ f u † ’ u ƒq x t† u‰ u d t  hq v˜¾…r‚iw8eC88×8eEsu8Ci&‡w™½€h¦&F‡iwxisC(‚wšigEssy¼&ijig‚sy¤‡sg‚F™‚syx Ý i) Å hulv „‘ u d iiy‚EšCFt ’ t m „ q u f … u d t f ’ v† x u ‰ ‰ f q t v u l ƒ u ‘ f m x ‰ h u l u “  q o f v ƒ p • x u ‘ uq m ¢ ’ t hq m „ f u v f u d t u q‰ w…‚¨€xC¤Ž‚d‚wÜ8wˆk‚€hsq`!”s‘wyE8–Ei˜w(€p‚E8‡e#‚€h¢%w‚…¯EEs‚žeFs–‚žwi6‚F¢€‘s˜ i) D h † Ÿ„ ‰ † ’ x h q t f q u ’ v m m f u t „ m f ’ ‘ ’ t f u v f q d t † ’ x u t f fq tx u u x „ ‰ à u s h d m f v  u d t h ‡8™sh}i¯C8s’w!(€f%y‚‚iBwC‚g8nwnwi¢€x‚FRijw!rsF–šw‚`sۉit’`‚iFišCF–8’ f uv f q dt u ‰t „ Ú h † Ÿ„ ‰ ’ t ‰ f „ Ù u q Ÿ ‘ … ƒ h f œ ‘ … h u u ot u • … Ó œ ‘ † …„  v u ƒ h „ f uv f u d t ’ widsx‚wš‚€hsqwC6¦‡i €h…FRi˜x3sxrØ×g‚!#ÖÁ©E“pET×v¤Ö¢2BC‚‚(wi#CF(˜ )  ‘ † …„ ‚ ut „ m f ’ˆžšwC‚g8’ Õ 1 2 0 x 3 ‘ †œ„ ‚ ut „ m f ¤–žšwC‚g8’ Õ ‘†… ’ˆ„ È ‚ A(x) 1 f(x)=1/x 2 £ œ f u‰ • ’ v m q w v ’ o v „ ’ p h ’ ƒ u x f 8¹Es‚iw‚–€h–w8“j‚888™yi¯Ô °†™iFv8!©‚Bu‚wBuE 3j!BCss`i”€‘FyE8¦‚w™‚!¤³… h f f  fq ƒ d t u ˜„ h … t f u hq‰ ‰ f q t v u l u d t ƒ h f œ ‘ u‚sh€q6‚Ft™E¨op‡•™v¤œ³ˆ2™8€’F‡‚‚ˆ6‚w…‚ij€xC¤Ž‚… ‰ u d h u u t u … Ó ‘ † …„  h qt ‘ h „† u dt ƒ h f q u f u d p • ƒ ƒ h „ • ‰ u dt u • ’t † …„ ‚ u h — CFtŒx‹u‚‚C8’‹‡Š¤ˆ‚‡ƒ‚F™tF#ˆ#ƒ‚˜‡u  8)  Ò Ì aS | Ð U Ï Î |S W Q | Í { Ì ¯bÑQ%2jkž¨E¯TµÀYŒË É “ f u‰ • ’ v m q † …„ ¶ ’ t ƒ u t f‰ u v † … ´ xq o ’ ÊsC8w‚–€h¯ˆ…w–Fv©y¯ˆ„ È ¢s”%3± É Ÿ f ‰ • ’ v m hq † …„ ª ’ t ƒ u t f‰ u v † … ¨ xq o ’ i€u‚8y‚s¨ˆ`#F–EF!sw¯„ È ¢s”%3± É œ f u‰ • ’ v m hq † …„  ’ t ƒ u t f‰ u v † … ‚ xq o ’ ¤¹Es‚iw‚s¨2rFw!©Ew¨ˆ„ È js”%3± ‘ † …„È ’ˆˆ¦´ ‘ † …„È ’ˆˆ¤¨ ‘ † …„È ’ˆ3‚ ² u l ’ • f “ ƒ h f Ÿ £ œ x f ‰ • ’ v m f ’ v† x h q t ‘ h „† f u v f u d t † ’ x u q t f q v u ƒ u d t u t f‰ „ ‰ f ‘ o ii¤Tin™‚inb8¦g€u‚8wCÀ8yˆ¢C8s’wE‚‚ˆžwiš‚w}i¢E8€lw!!€ly‚B‚wdF!s‚€‘i”Ǥ’ Æ Å ”' ›†EFv©¢!˜w–F€t¢r!”‚—€‰Ä‰!½2C™i‚F¯iwisqwE8gƒE‚¢“ˆrgw€t¢rp€t‚Erc‚ƒ h v u t f‰ t f d t d q o ‰ f u ƒ à u s h ¸ h f d t v u  v f‰ x t u  … h u d o † …„ ´ d q o p ‰ „ ‘ q f–€xÁwE‚j¡À‚‚€hwiwE‚…€x(ˆ¿‚iz‚€hwiwE‚shrsxrr“”i&–‚8E…s(2¿8‚8CF¾8‡…‚8€’F‡‚‚ˆ† q u v u d i„ h  q x f u v ‘ u ƒ q † …„ ¶ ƒ h f  q x f u v ‘ q q † …„ ª £ t h f tx h ’ ‘ xq † …„  d  „ ’ d t h u l u x h q t ‘ h „  qx f uv ‘ q uv f † …„ ´ ƒ h f † …„ ¨ £ † …„ ‚ ’ ˜ h x ux f uv ‘ q … x f x ux f uv ‘ q ‡ ‰ ˆ ‡ u d t £ u l ’ • f u d t † ’ d ‘ f u v ’ Cshw!”wE‚shnyi¦ˆr½‚!B–‡#(EwiwCsh¼¦i¤yi”y‚€htx¤‚g‚Fr!8%Tig‚F“it&!”¦8˜» É v ’ t f‰ „ ‘‰ f &8w!©Cs!”‘ v „ ’ p p • h u q  u ‰ f l u dt ’t v u ox h f tx ux ‰ ‘ u dt „ ’ p x u q  x ut f qtx u v „ ’† v „ ’ p † ’ d q d s É ut f qtx ‚i8©x½8cl8g‚s„i!(CFƒwž“y‚!#yEw8€’…CF¼8i¤i€l8¤w!(€fF–š‚8ˆe‚i8ji¼&€‘‚¾ã%w!(€fF–u v u ƒ h „ ƒ h f v ’ u t f q t x uv u l ’ h f  q t t u  u v f „ ’ p q ‰ u t „ ’ p h h † Ÿ„ h‰ u t f q u ’ v m m f ’ t f u d t u Ž‚C‚z‚i#8(F!rsfwyP–E8%¦iׂ€hFw‡8…yiŒ88¢€†ks‰EF¦8iz!f Õ â8 rs…w!(€f%y‚‚i×wåE‚F…yx Ý h x ‰  h f t ‘ u v ¢  q ‰ „ x u v u d t q d ‘ t u w x ƒ h f t d q u d u d t u q f v u t u ƒ ’ t u ‘ ‰ x d ‘ f u † ’ t hq ’ m q f u d t u €u8‚iFEEwB6‚€hF€tCwyj‚Fž€h(&wEiyn‚i’8€‚j‚w¨‚€hryF‡‚BFks€qwg&i”Ti}s8T‚€ƒrj‚wjwx Ý i) ä É v ’ t f‰ „ ‘‰ f ‘ v „ ’ p p • h u q  u „‰ f l u d t d q &8w!©Cs!”q‚88(x™i€l8¦Csi!¦‚F–w€t¢o u v f m f ’ ‘ q x u ’ ƒ o ’ ± h † Ÿ„ ‰ † ’ u t f q t x u o u h f t u  ’ t x u t f fq t x u ’ ot v „ ’ p † ’ u  f v u l f u d t u w f á yi˜g8d€t6ECã%3Bâ8 rsh’ieF!rsfwy³‡‚…BEi–F3EF!rswyE™“pd‚i8’i#8iwi%ž‚Fž!iYi !) É u t f q t x uv u ƒ h „ h f v ’ u t f fq tx uv u l ’ h f q xà h † Ÿ„ h‰ ƒ h f £ u l v „ ”F!rsfwyEp–E‚‚‚™i¢8¦w!rsF–P–i%ƒiqct¢p‡8™—s™‚iT!8yC‘ uCF¨E‚‚‚ewid‚F`iris’Fv(€fC”y‚‚i‚iy¨‚i8’sx¨€x‚jr‚ÖCFrCi–‚i¨€u8‚iFEy6wE‚F`ie”yif dt vu ƒ h „ fuvf u dt †’ h qtf q u’v m mf ƒ h’‘ux v „’ p q q d i h m „ fu dt ƒ ƒf ƒ hf x ‰  hft‘uv uxu dt †’ fuv u d t ƒ q » h d m f v  v „ ’ p h ’ x ‰  h f t ‘ u v ¢ q d ‘ t u w x £ t d q u d u d t x f u ‘ ‰ x d ‘ f u † ’ u ƒq x t d q v u d t  hq CFrCshY`‚iFi¯‚8ir8“sui‚i&‡ye(€h&FE!wŠ”8€‚3‚F’i3s€qw–&i—iq‚sy¨8€w3‚Fš‚syx Ý i) £ Antiderivatives and Areas Math 125 Name Quiz Section In this worksheet, we explore the Fundamental Theorem of Calculus and applications of the Area Problem to problems involving distance and velocity. We also consider integrals involving net and total change. FTC Practice 1 Let f (x) be given by the graph to the right and define 3 x A(x) = y=f(x) f (t) dt. Compute the following. 0 2 A(1) = A(2) = A(3) = A(4) = A (1) = A (2) = A (3) = A (4) = 1 1 2 4 3 5 The maximum value of A(x) on the interval [0, 5] is The maximum value of A (x) on the interval [0, 5] is Velocity and Distance 2 A toy car is travelling on a straight track. Its velocity v (t), in m/sec, be given by the graph to the right. Define s(t) to be the position of the car in meters. Choose coordinates so that s(0) = 0. Compute the following. 3 2 y=v(t) 1 s(2) = s(4) = s(6) = 1 v (2) = v (4) = v (6) = The maximum value of s(t) on the interval [0, 7] is The minimum value of s(t) on the interval [0, 7] is The maximum value of v (t) on the interval [0, 7] is The minimum value of v (t) on the interval [0, 7] is -1 2 3 4 5 6 7 Net and Total Change 2 3 2 x2 − 4 dx and (a) Evaluate −2 x2 − 4 dx and explain your answers. −2 3 3 x2 − 4 dx and explain your answers. x2 − 4 dx and (b) Now evaluate −3 −3 0 2 4 6 ¦ ¥£ ¢ £¢ ¢ ¢ ¢ ¢ ¢£¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦£ ¤¤£¤£¤£¤£¤£¤£¤£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£ ¥£ ¢£¤£¤£¤£¤£¤£¤£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£ £¢£¢£¢£¢£¢£¢£¤ ¡¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥¦ ¦£ £¤£¤£¤£¤£¤£¤£¢¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£ ¢¤£¢£¢£¢£¢£¢£¢£¤¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥¦ ¦¥£ £¤£¤£¤£¤£¤£¤£¢¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£ ¥£ ¢¤£¢£¢£¢£¢£¢£¢£¤¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥¦ ¦£ £¤£¤£¤£¤£¤£¤£¢¤¢¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¥¦ ¥£ ¦£ ¥£ ¢¢¤£¢£¢£¢£¢£¢£¢£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£ £¢£¢£¢£¢£¢£¢£¤¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¥¦ ¦£ ¤¤£¤£¤£¤£¤£¤£¤£¢¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£ ¥£ ¢£¤£¤£¤£¤£¤£¤£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£ £¢£¢£¢£¢£¢£¢£¤¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥¦ ¦£ £¤£¤£¤£¤£¤£¤£¢¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£ ¥£ ¢¤£¢£¢£¢£¢£¢£¢£¤¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥¦ ¦£ £¤£¤£¤£¤£¤£¤£¢¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£ ¥£ ¢¤£¢£¢£¢£¢£¢£¢£¤¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥¦ ¦£ £¤£¤£¤£¤£¤£¤£¢¤¢¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¥¦ ¥£ ¦£ ¥£ ¢¢¤£¢£¢£¢£¢£¢£¢£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£ £¢£¢£¢£¢£¢£¢£¤££££££££££££££££¥¦ ¦£ ¤¤£¤£¤£¤£¤£¤£¤£¢¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¦ ¥£ ¢£¤£¤£¤£¤£¤£¤£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£ £¢£¢£¢£¢£¢£¢£¤¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¥¦ £¤£¤£¤£¤£¤£¤£¢¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£ ¢¤£¢£¢£¢£¢£¢£¢£¢¤¥¦£¥¦£¥¦£¥¦£¥¦£¥¦£¥¦£¥¦£¥¦£¥¦£¥¦£¥¦£¥¦£¥¦£¥¦£¥¦£¥¦ ¦¥¦£ ¢¤£¢¤£¢¤£¢¤£¢¤£¢¤£¢¤£¢¤¥¦£¥¦£¥¦£¥¦£¥¦£¥¦£¥¦£¥¦£¥¦£¥¦£¥¦£¥¦£¥¦£¥¦£¥¦£¥¦£¥¦ ¢¤£¢¤£¢¤£¢¤£¢¤£¢¤£¤¥¦£¥¦£¥¦£¥¦£¥¦£¥¦£¥¦£¥¦£¥¦£¥¦£¥¦£¥¦£¥¦£¥¦£¥¦£¥¦£¥¦ ¦¦¥¦¥ ¤£££££££¢¤¢¥¦£¥¦£¥¦£¥¦£¥¦£¥¦£¥¦£¥¦£¥¦£¥¦£¥¦£¥¦£¥¦£¥¦£¥¦£¥¦£¥¦ ¢¢¤£¤¤¢£¤¤¢£¤¤¢£¤¤¢£¤¤¢£¤¤¢£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£ ¢¤¤£¢¤£¢¤£¢¤£¢¤£¢¤£¢¤£¥¦£¥¦£¥¦£¥¦£¥¦£¥¦£¥¦£¥¦£¥¦£¥¦£¥¦£¥¦£¥¦£¥¦£¥¦£¥¦£ £¤£¤£¤£¤£¤£¤£¤¢££££££££££££££££¥¦ ¦¥£ ¥£ ¢¢¤£¤£¤£¤£¤£¤£¤£¢¤¤¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥¥¦¦ ¦¥£ £¢£¢£¢£¢£¢£¢£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£ ¥£ £¢£¢£¢£¢£¢£¢£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£ ¦£ £¢£¢£¢£¢£¢£¢£¢¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£¦£ ¥£ £ ¢¤£¤£¤£¤£¤£¤£¤£¤¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥£¥¦ £¢£¢£¢£¢£¢£¢£¤¢££££££££££££££££¥¦ ¤¢¤£¤£¤£¤£¤£¤£¤£¢ ¢£¢£¢£¢£¢£¢£¢£ £¤£¤£¤£¤£¤£¤£¤ £¢£¢£¢£¢£¢£¢£¢ ¢¤£¤£¤£¤£¤£¤£¤£¤ £¢£¢£¢£¢£¢£¢£¢ ¢¤£¤£¤£¤£¤£¤£¤£¤ £¢£¢£¢£¢£¢£¢£¢ ¢¤£¤£¤£¤£¤£¤£¤£¤ £¢£¢£¢£¢£¢£¢£¤¢ ¤¢¤£¤£¤£¤£¤£¤£¤£¢ ¢£¢£¢£¢£¢£¢£¢£ £¤£¤£¤£¤£¤£¤£¤ £¢£¢£¢£¢£¢£¢£¢ ¢¤£¤£¤£¤£¤£¤£¤£¤ £¢£¢£¢£¢£¢£¢£¢ ¢¤£¤£¤£¤£¤£¤£¤£¤ £¢£¢£¢£¢£¢£¢£¢ ¢¤£¤£¤£¤£¤£¤£¤£¤ £¢£¢£¢£¢£¢£¢£¤¢ ¤¢¤£¤£¤£¤£¤£¤£¤£¢ ¢£¢£¢£¢£¢£¢£¢£ £¤£¤£¤£¤£¤£¤£¤ £¢£¢£¢£¢£¢£¢£¢ ¢¤£¢¤£¢¤£¢¤£¢¤£¢¤£¢¤£¢¤ ¢¤£¢¤£¢¤£¢¤£¢¤£¢¤£¢¤£¤¢¤ ¤¢¤£¢¤£¢¤£¢¤£¢¤£¢¤£¢¤£¢¤¢ ¢¤£¢¤£¢¤£¢¤£¢¤£¢¤£¢¤£¢¤ ¤¢£¤£¤£¤£¤£¤£¤£ ¢£¢£¢£¢£¢£¢£¢£ £££££££¢¢¤¤ (ii) Where should the artist divide the region with vertical lines to get 3 pieces with equal areas? (i) Where should the artist divide the region with a vertical line so that each piece has the same area? (See the picture.) 4 An artist you know wants to make a figure consisting of the region between the curve y = x2 and the x-axis for 0≤x≤3 1 2 ? 3 y=x2 8 10 Another Area Problem Area Between Curves Math 125 Name Quiz Section In this work sheet we’ll study the problem of finding the area of a region bounded by curves. We’ll first estimate an area given numerical information. The we’ll use calculus to find the area of a more complicated region. The Lake 70 feet 80 feet 50 feet 45 feet 65 feet 1 The widths, in feet, of a small lake were measured at 40 foot intervals. Estimate the area of the lake. 40 feet Area Bounded by Three Curves 2 On the grid below sketch the graphs of y = 4, y = x2 and y = piece of a sideways parabola). √ 27x. (The last one is just a 16 12 8 4 0 1 2 3 4 3 Shade the “triangular” region bounded by the graphs of the three functions that lies above the horizontal line. 4 Compute the x-coordinate of the left endpoint of the region. 5 Compute the x-coordinate of the right endpoint of the region. 6 Note that the top of the region consistsof a singlecurve, but the bottom of the region consists of two different curves. Find the x-coordinate where thesetwo curves meet. 7 Sketch in a vertical line at the x-coordinate you found in the last problem. This divides the region into two smaller sub-regions. 8 Compute the area of the left sub-region. 9 Compute the area of the right sub-region. Add the two areas together to get the total area. 10 Recompute the area using the following trick. Solve for x as a function of y in the two non-constant functions. Find the area by integrating with respect to y . Is this easier? Solids of Revolution and the Astroid Math 125 Name Quiz Section In this worksheet we are going to practice computing some more volumes of solids of revolution. These will all be based on a curve called the “astroid”. This curve is formed by rolling a small wheel around the inside of a larger one (see the picture). If the radius of the small wheel is one quarter the radius of the big one, a point P on the small wheel will trace out the four pointed curve shown on the far right. It’s called the astroid because it looks like a star. 1 If the radius of the big wheel is taken to be one, the astroid can be shown to have the equation x2/3 + y 2/3 = 1. Use disks to compute the volume of the solid generated by rotating the part of the astroid in the first quadrant around the y -axis. P 1 0 1 2 Use cylindrical shells to compute the volume of the solid generated by rotating the first quadrant portion of the astroid about the x-axis. (Hint: Try the substitution u3 = y 2 , so 3u2 du = 2y dy .) How does this compare with your answer in Problem 1? Can you explain this geometrically? 3 Use any method you wish to compute the volumes of the solids generated by rotating the first quadrant portion of the astroid about the lines x = 1 and y = −1. Set up only. Do not compute the integrals. Integration by Parts Math 125 Name Quiz Section In this work sheet we’ll study the technique of integration by parts. Recall that the basic formula looks like this: u dv = u · v − 1 First a warm-up problem. Consider the integral v du x sin(3x) dx. Let u = x and let dv = sin(3x) dx. Compute du by differentiating and v by integrating, and use the basic formula to compute the original integral. Don’t forget the arbitrary constant! 2 Compute ln x dx. (The proper technique is, indeed, integration by parts. What should you take to be u and dv ? The choices are pretty limited. Try one and see what happens.) (c) Use the shell method to find the volume of the solid swept out by region B . (b) Use the disk method to find the volume of the solid swept out by region A. 1 2 e3 4 5 ¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¤¡¤¡¤¡¤¡¤¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡¢£¡£¡£¡£¡£¡¤ ¤£¤ ¢¤£¤ ¢¤£¤ ¢¤£¤ ¢¤£¤ ¢¤£¤ ¢¤£¤ ¢¤£¤ ¢¤£¤ £¤ £¤ £¤ £¤ £¤ ¢¡¡¡¡¡¡¡¡¡¢ ¡¡¡¡¡¤£ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ £¡£¡£¡£¡£¡£ ¤£ ¤£ ¤£ ¤£ ¤£ ¤£ ¤£ ¤£ ¤£ ¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¤¡¤¡¤¡¤¡¤¡ £¡£¡£¡£¡£¡£¡£¡£¡£¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡¢£¡£¡£¡£¡£¡¤ ¤¡¢¤¡¢¤¡¢¤¡¢¤¡¢¤¡¢¤¡¢¤¡¢¤¡¤ ¤ ¤ ¤ ¤ ¢¡¡¡¡¡¡¡¡¡¢ ¡¡¡¡¡¤£ £¡£¡£¡£¡£¡£¡£¡£¡£¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ £¡£¡£¡£¡£¡£ ¤¡¤¡¤¡¤¡¤¡¤¡¡¤¡¤¡ ¢¡¡¡¡¡¡¡¡¡¡¡¡¡¡ £¡£¡£¡£¡£¡£¡£¡£¡£¡ ¢¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¢£¡£¡£¡£¡£¡£¤ ¤¡¢¤¡¢¤¡¢¤¡¢¤¡¢¤¡¢¤¡¢¤¡¢¤¡¤ ¤ ¤ ¤ ¤ £¡£¡£¡£¡£¡£¡£¡£¡£¡ ¢¡¡¡¡¡¡¡¡¡¢ ¢¡¡¡¡¡¤£¤ ¤£¡¤£¡¤£¡¤£¡¤£¡¤£¡¤£¡¤£¡¤£¡ ¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¤ ¤ ¤ ¤ ¤ ¡¢ ¡¢ ¡¢ ¡¢ ¡¢ ¡¢ ¡¢ ¡¢ ¡¤£¡¤£¡¤£¡¤£¡¤£¡ ¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡ ¤¡¤¡¤¡¤¡¤¡£ ¤¡¤¡¤¡¤¡¤¡¤¡¤¡¤¡¤¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡£¡£¡£¡£¡£¡ ¡¡¡¡¡¡¡¡¡ ¢¡¡¡¡¡£¤ £¡£¡£¡£¡£¡£¡£¡£¡£¡ ¡¡¡¡¡¡¡¡¡ A (a) Before computing the integrals, which solid do you think has a larger volume? Why? 1 B y=ln(x) 3 The regions A and B in the figure are revolved around the x-axis to form two solids of revolution. 4 Suppose we try to integrate 1/x by parts, taking u = 1/x and dv = dx. We have du = (−1/x2 ) dx and v = x, so 1 −1 1 dx = · x − x · 2 dx x x x 1 = 1+ dx. x Canceling the integral from both sides, we get the disconcerting result that 0 = 1. What went wrong? What happens if we replace the indefinite integrals by definite integrals, that is, if we try b1 dx by this method? to calculate ax Algebra and Partial Fractions Math 125 Name Quiz Section Integration of rational functions is mostly a matter of algebraic manipulation. In this worksheet we shall work through some examples of the necessary techniques. 2x3 − 4x2 − 5x + 3 . Use long division to get a quotient x2 − 2x − 3 and a remainder, then write f (x) =quotient + (remainder/divisor). 1a Consider the rational function f (x) = 1b Now consider the expression x2 x+3 . Factor the denominator into two linear terms. − 2x − 3 A B x+3 as a sum + . Let’s find A and B . Set the two (x − 3)(x + 1) x−3 x+1 expressions equal and clear denominators (that is, multiply through by (x − 3)(x + 1) and cancel (x − 3)’s and (x + 1)’s as much as possible). Plug in x = 3 and solve for A. Use the same idea to find B . Check your work by adding the two fractions together. 1c We wish to write 1d Now use the results of Problems 1a, 1b, and 1c to compute 2x3 − 4x2 − 5x + 3 dx. x2 − 2x − 3 1e Some of the terms in the answer to Problem 1d involve logarithms. Combine those terms into a single term of the form ln(some function of x). 3x + 1 . The problem here is that one of the linear x(x + 1)2 factors in the denominator is squared. Partial fraction theory says the best we can do is to get this A B C one in the form + + . Let’s find A, B and C . x x + 1 (x + 1)2 2a Next, consider the function f (x) = First set the two expressions equal and clear denominators. Plug in x = 0 and solve for A. Plug in x = −1 and solve for C . 2b Now that you’ve found A and C , you can find B by plugging in any other convenient value for x. Do so. 2c Now compute 3x + 1 dx. x(x + 1)2 Techniques of Integration Name Math 125 Quiz Section The following integrals are more challenging than the basic ones we’ve seen in the textbook so far. You will probably have to use more than one technique to solve them. 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Forensic Mathematics A detective discovers a murder victim in a hotel room at 9:00am one morning. The temperature of the body is 80.0◦ F. One hour later, at 10:00am, the body has cooled to 75.0◦ F. The room is kept at a constant temperature of 70.0◦ F. Assume that the victim had a normal temperature of 98.6◦ F at the time of death. We’ll use differential equations to find the time the murder took place. Let u(t) be the temperature of the body after t hours. By Newton’s Law of Cooling we have the differential equation du = k (u − 70) dt where k is a constant (to be determined). We’ll solve the differential equation and get a formula for u(t). 1 Multiply both sides by dt to get a differential form of the equation. Now do some easy algebra to get the variable u on the same side as the du. (Leave the k where it is). 2 Integrate both sides of the equation. Integrate the right side with respect to t and the left with respect to u. You can combine the integration constants into one “+C ” on the right side. 3 Solve for u as a function of t. Your function will involve the constants k and C . 4 Take t = 0 when the body was found at 9:00am. Plug in t = 0 and u = 80.0◦ F and solve for C . (It’s easier to solve for A = eC and use this in your formula). 5 Plug in t = 1 and u = 75.0◦ F and solve for k . (This’ll take some log tricks). 6 Set u = 98.6◦ F and solve for t. At what time did the murder take place? Spread of a Rumor The Xylocom Company has 1000 employees. On Monday a rumor began to spread among them that the CEO had suddenly moved to Brazil. It is reasonable to assume that the rate of the spread of the rumor is proportional to the number of possible encounters between employees who have heard the rumor and those who have not. Let y = y (t) be number of employees who have heard the rumor after t days. 1 Explain why the number of possible meetings between employees who have heard the rumor and those who have not equals y (1000 − y ). 2 Write a differential equation that describes this model of the spread of a rumor. (Remember that “is proportional to” means “is some constant k times”.) 3 Proceed as in the cooling body problem to solve the differential equation for y (t). You will need to use the method of partial fractions. Your answer should involve two constants: the proportionality constant k and a constant C from integrating. (As in part 2 of the previous problem, you can combine the integration constants into one “+C ” on the right side.) 4 At the very beginning, 50 people had heard the rumor (they all attended the same meeting). Compute the constant A = eC . 5 On Tuesday morning, 100 people had heard the rumor. 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¢¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢ ¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡¢ ¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡¢ ¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡¢ ¢ ¢¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡ ¢¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡¢ ¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡¢ ¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡¢ ¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¢ ¢ ¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡ ¢ ¢ 1.5 2. Find the center of mass of a plate with constant density 1 that occupies the region − 4 π ≤ x ≤ 0, 0 ≤ y ≤ sec2 x. 2 Center of Mass 1. Write an integral that computes the arclength of the curve y = ex/2 between x = 0 and x = 2. Use Simpson’s Rule with n = 4 subintervals to estimate the value of the integral. Arclength and Approximation The following problems should help you review for the final exam. Don’t hesitate to ask for hints if you get stuck. Name Quiz Section Review for the Final Exam - Part 2 Math 125 Net and Total Distance 3. You throw a ball straight up into the air with velocity 40 ft/sec and catch it (at the same height) when it comes back down. What is the total distance traveled by the ball? Differential Equations t Let f (t) be a continuous function and let a be a constant. Show that y = e−at dy + ay = f (t). satisfies the differential equation dt eas f (s) ds 4. 0 5. An electric circuit with resistance 10 ohms and inductance 2 henrys is powered by a 12-volt battery. The current I (in amperes) at time t (in seconds) in such a circuit satisfies the differential equation dI 2 + 10I = 12. dt Suppose that I = 0 when the circuit is activated at time t = 0. (a) Find the current I at all times t > 0. (b) Find the limiting value of I as t → ∞. (c) After what time is the current within 0.1 ampere of its limiting value? ...
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This note was uploaded on 10/24/2011 for the course MATH 125 taught by Professor Chen during the Spring '08 term at University of Washington.

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