HW1 - MATH 191 (Spring 2008) page 1 Professor: Dennis Yang...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: MATH 191 (Spring 2008) page 1 Professor: Dennis Yang TA: Kwang Taik Kim Since f is increasing on [0, 1], we use left endpoints to obtain lower sums and light endpoints to obtain upper sums. + 1>x 1 i 1 . (a) Ax:1&9=%andxi:iAx=§=>alowersumi32( )3'%=%(03+(%)3>:i NI- . 16 1:0 . 3 , <b> Ax: 1—269=iandxa=iAx=iealowersumisflifi=%(03+(%)3+(%)3+(%)3) = = i=0 . 2 p i (C) AX : 1%] z and xi 2 iAX : é :2 an uppersumis - % = %<(%)3+13) : % . g _—_ % i=1 __ , i , 4 i (d) Ax = 1—3 = g and xi : 1Ax= i :> anuppersum1s2(fi)3 . i I “(g3 + (35 +(§)3+13) = 2 % 2 3.: i=1 5/ Since f is increasing on [~2, 0] and decreasmg on [0, 2], we use left endpoints on [—2, 0] and right endpoints on [0, 2] to obtain lower sums and use right endpoints on [—2, 0] and left endpoints on [0, 2] to obtain upper sums. » -2 2 (a) Ax: 31%12122andxi=—2+iAx=—-——2+2i=>alowersumis2- (4—(~2)2)+2-(4—22)=0 1. 4 (b) Ax = i441 = 1 andxi = ~2+i_Ax= —-2 +i => alowersum is 2(4— (x02) ~ 1 + 2(4 ——— (x02) o 1 i=0 i=3 = 1((4 " (4)2) + (4 —(-.1)2)+(4 - 12) + (4 — 22)) = 5 I (c) Ax: bylaw“ =‘2+iAX-= ~2+21i=>auppersm‘nis?- (4~ (0)2) +2~(4-—02) == 16 . p 2 3 p (d) AX = ~42"? = 1 andxi = ~2+iAx = ~2 +i=> aupper sum is Z(4—— (x02) . 1 + T (X02) . 1 i=1 i=2 = 1((4"(4)2)+9432)+(4§02)+(4~- 12)) = 14 10. (a) D R: (1)600) + (1.2)(300) + (1.7)(300) + (2.0)(300) + (1:8)(300) + (1.6)(300) + (1.4)(300) + (1.2)(300) + (1.0)(300) + (1.8)(300) + (1.5)(300) + (1.2)(300) = 5220 meters (NOTE: 5 minutes 2: 300 seconds) (b) D m (1.2)(300) + (1.7)(300) + (2.0)(300) + (1:8)(300) (1.6.)(300) + (1.4)(300) + (1.2)(300) + (1.0)(300) ' + (1.8)(300) + (1.5)(300) + (3.2)(300) + (0)300) = 4928 meters (NOTE: 5 minutes =: 300 seconds) MATH 191 (Spring 2008) page 2 Professor: Dennis Yang TA: Kwang Talk Kim ii 14. (a) The speed is a decreasing function of time :> right end—points give an lower estimate for the height (distance) attained. Also (away) t 0 l 2 3 4 5 V 400 368 336 304 272 240 gives the time—velocity table by subtracting the constant g = 32 from the speed at each time increment At = 1 sec. Thus, the speed z 240 ft/sec after 5 seconds. We,» gg-L‘M—ie 1‘— Dfl). M/gaa (b) A lower estimate for height attained is h m [368 + 336 + 304 + 272 + 240]( 1) = 1520 ft. Section 5.2 Sigma Notation and Limits of Finite Sums 6. i(—l)k coskir = (—1)1 cos(lzz)+(—1)2 COS(27Z')+(—l)3 cos(37£)+(—l)4 cos(47:) =—(—1)+1—(—1)+1 =4 Since f is increasing on [0, l] we use right endpoints to obtain upper sumslkx 2 1—;9 2 i and xi 2 iAX EWH upper sum . n n - 2 1. IS 213%) = we (a = 12 = - (——"<“+1>ge—w) 1: i=1 i:l 3 1 n ._ gns+3n2+n _ 2+E+32 - 2 1 — 2H3 — ' 2 ~. Thus, 11m 23xi (a) Ill—’00 3 1 =lim “5+2 : {Ev->00 2 :1, MIN ...
View Full Document

This note was uploaded on 06/19/2008 for the course MATH 1910 taught by Professor Berman during the Spring '07 term at Cornell University (Engineering School).

Page1 / 2

HW1 - MATH 191 (Spring 2008) page 1 Professor: Dennis Yang...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online