# Exam 1 - Version 105 w EXAM 1 M hamrick w (57395) This...

This preview shows pages 1–14. Sign up to view the full content.

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

View Full Document

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

View Full Document

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

View Full Document

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

View Full Document

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

View Full Document

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

View Full Document

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Version 105 w EXAM 1 M hamrick w (57395) This print~out shouid have 17 questions. Multiple-choice questions may continue on the next column or page W find all choices before answering. .. '1 '. E 5“ 4 1-. 001 10.0 points If the graph of f is which one of the following contains only graphs of anti-derivatives of j"? 002 10.0 points Find f(1)w11e11 f”(:r:) 2: 6:1: and f(-~1) = 5.. f’(—1) m 5. M H} 2‘“ pm: ‘-_/ ii H yﬁ. Version 105 ~ EXAM 1 w hamriek —- (5?395) 2 3. f(1) = 13 005 10.0 points 4. fa) : 11 Use properties of integrals to determine the value of 0 5. f(1) z: 12 I m1} fwd-T when 003 10.0 points 6 6 Estimate the area, A, under the graph of / dm 2 3, / f(;t-)dzc m . o . 4 4 :r = m f( ) a: L I a 5 on [1, 5] by dividing [L 5] into four equai subintervais and using right endpoints. 2 I m w2 76 1. A z 3_ I z 4 77 2. A z ~— 15 4. I z “4 3. A z 5 5. I : W5 26 4. A a w» 5 6 I 2 74 ‘ x 5. A m —— 15 084 10.0 points 006 10.0 points The graph of f is shown in the figure Evaluate the deﬁnite integrai I = [1 (1+\/16w:172) dam .m: by interpreting it in terms of known areas. 1. I w 167r—4 81r+8 e 3. I :167T+8 4, I m 167T + 4 If F is; an anti-derivative off and 3 ( 5.1:8Trm—4 /f(:r:)d;r m 1 6. I = 87:“ + 4 ﬁnd the value of F(8) —— F(0). Version 105 — EXAM 1 — hamrick — (57395) 3 '"1 1 F(8) Mme) = 3g 105 3. F(8)WF(0) 2 27 . 009 10.0 points ' F(8)-F(0) \$24 Evaluate the deﬁnite integral 99 5 F 8 -F 0 w — ﬁ/éi ( ( ) ( ) 4 In": / {Ssecgmu—Sseegmtanzx}dm. - 0 007 10.0 points 1. I 8 Evaluate the definite integral I = /4 (llx2wélwm2)dm. 0 13 1.1: m 3 2.1: 3:: 23 5.1 : , 10 5 010 10.0 points Deterl'niue the indeﬁnite integral 008 10.0 points Evaluate the integral 1 I m/ 5m(1——:I:2)5dm. 0 5 1.1:” 6 2.12mi 1 3.I——-2- Version 105 w EXAM 1 w hamrick — (57395) 4 011 10.0 points 3. F’(:27) m 6m/1+m3 Find the value off(—-1)When 4. Pkg) 3 3\$2W f’(:c) : 2x849, m) m 7 6 5. F’ a: :2 ——\$——-— 1 7 ( ) V1+Ji3 MWW‘WJ'“""‘“"Z”“”“""'"”ww 014 10.0 points 3 f(_l) : -384? Find the area of izhe region enclosed by the 7 graphs of 1 n . 4. f(——1) = ?e‘m§;{g = 4+mwm2, 9(1)) : .1:+1, 50 1 w. ontheintervai[0,1}. 5- f(m1) w - :e” 7 f 8 2 7 sq. units *0 1 ~ 3 6 j'(—1) 2: 3mwwe* 7 7 10 2. Area *2 —3—sq. units 012 10.0 points If the graph of f passes through the point 3. Area 2 3 sq. units (1, MS) and the siopo of the taingoni line at (at, f(:(:)) is 4:17 W 9, find the value of f(2). 4. Area 2 25g. units 1.f(2)= m7 2x “Mic 7 0/” _ .. \ . c: 5. Area 2 —— sq. units /” c 9" 1- t: _ v»- ai. "5 if ‘ I; /2./ j-‘(2)W\“~~——w ~87 is” J" 3 NM / r ‘3. , EWWWMM "w 015 10.0 oints 3- f(2) : “5 {CUO "' QMZJiXi—E’ p 7‘ The shaded region in the figure 4. .f(2) m “9 H25): {ii—m. H? w. 24-“5 5. f(2) = “6 013 10.0 points Determine F’(:1:) when m2 = / 3V1+t3df M ’5 1. Fr .) = MM 7’" 3 9 [0 ‘L m“ is bounded by the graph of 2. F’(:r-) = m \/1 + :1?“ :1: m 63; — yg Version 105 — EXAM 1 w harnriek ~~ (57395) 5 and the y»axie. Find the area of this region. Wm sq.units 37 sqenits 109 1. A z: rea 3 2. Area r» 11 3. Area z {‘30 1 4. Area m M»? /”’MMWMM¢MKH\V / K squnits sq.units 5. Area we» its wwww 016 10.0 points The shaded region shown in is enclosed by the graphs of yew, 2 army2 1‘; TV»: Find the voiume, V, of the scﬁid obtained by rotatingwtvhga shaded region about the 1:»axis. / 917 10.0 points A stone is dropped off a chi? and falls un— der. gravity With a constant acceieration of ~32 ft/sec2. If it hits the ground with a speed of 128 ft/sec, determine the height of the cliﬁ. 1. height = 264 ft 2. height 2 252 ft 3. height H 260 ft x, “m” __ {“UQ = r -~ 7-3 157 C FCTQ”‘Q¥%: 2 M5“ 1;!(4) :: 36'!) “M: W a”: / 3+C.= :73"? 3 xzﬁ-i-Z \_ 3 ‘ c 15021”: 3’4 "le‘m 9.. g! __I {if}. M‘ 4‘" C. i: 6 «CLbe ‘ ‘7 "L " wf__2_-§"C.m\$ «‘3 m ..r.. :1. w I'" win (:1 __m ’ \Jr "ng/k) W. X % «z m f<1+ ma)g€.>< a»; n ...
View Full Document

## This note was uploaded on 03/24/2010 for the course M 57395 taught by Professor Hamrick during the Fall '09 term at University of Texas.

### Page1 / 14

Exam 1 - Version 105 w EXAM 1 M hamrick w (57395) This...

This preview shows document pages 1 - 14. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online