Test 3 Review. Jiwen He. I will replace your lowest test score with the percentage grade from the final exam (provided it is higher).


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1 Test 3 Review Jiwen He Test 3 Test 3: Dec. 46 in CASA Mteril  Through 6.3. No Homework (Thnksgiving) No homework this week! Hve GREAT Thnksgiving! Finl Exm Finl Exm: Dec in CASA You Might Be Interested to Know... I will replce your lowest test score with the percentge grde from the finl exm (provided it is higher). I will give n A to nyone who receives 95% or bove on the finl exm. I will give pssing grde to nyone who receives t lest 70% on the finl exm. Quiz 1 Wht is tody?. Mondy b. Wednesdy c. Fridy d. None of these 1
2 Test 3 Mteril Test 3 will cover mteril from Chpter 5, long with Sections 6.1, 6.2 nd 6.3. Good Sources of Prctice Problems Exmples from clss. The bsic homework problems. The bsic online quiz problems. Definite Integrl nd Lower/Upper Sums Are of Ω = Are of Ω 1 + Are of Ω Are of Ω n, L f (P ) = m 1 x 1 + m 2 x m n x n U f (P ) = M 1 x 1 + M 2 x M n x n L f (P ) f(x) dx U f (P ), for ll prtitions P of [, b] Problem 1 Give both the upper nd lower Riemnn sums for the function f(x) = x 2 over the intervl [1, 3] with respect to the prtition P = {1, 3 2, 2, 3}. 2
3 Lower/Upper Sums nd Riemnn Sums S (P ) = f(x 1) x 1 + f(x 2) x f(x n) x n L f (P ) S (P ) U f (P ), for ll prtitions P of [, b] Problem 2 Give the Riemnn sums for the function f(x) = x 2 over the intervl [1, 3] with respect to the prtition P = {1, 3 2, 2, 3} using midpoints. 3
4 Fundmentl Theorem of Integrl Clculus Theorem 1. In generl, where F (x) is n ntiderivtive of f(x). f(x) dx = F (b) F (). Problem 3 4
5 Evlute π/4 0 ( 2x 6x ) dx x x 2 dx ( 4 x ) 2 dx sec x ( 2 tn x 5 sec x ) dx Are below the grph of Nonnegtive f f(x) 0 for ll x in [, b]. Ω = region below the grph of f. 5
6 Are of Ω = where F (x) is n ntiderivtive of f(x). Problem 4 f(x) dx = F (b) F () Find the re bounded bove by the grph of f(x) = x 2 nd below by the xxis over the intervl [1, 3]. Are between the grphs of f nd g f(x) g(x) for ll x in [, b]. Ω = region between the grphs of f (Top) nd g (Bottom). Are of Ω = [ ] b [ ] Top Bottom dx = f(x) g(x) dx. 6
7 Problem 5 Find the re between the grphs of y = 4x nd y = x 3 over the intervl [ 2, 2]. f(x) dx s Signed Are f(x) dx = c f(x) dx + d c f(x) dx + e d f(x) dx + e f(x) dx = Are of Ω 1 Are of Ω 2 + Are of Ω 3 Are of Ω 4 = [ Are of Ω 1 + Are of Ω 3 ] [ Are of Ω2 + Are of Ω 4 ] = Are bove the xxis Are below the xxis. Problem 6 The grph of y = f(x) is shown below. The region Ω 2 hs re 3 nd is 2. Give the re of region Ω 1. c f(x)dx 7
8 Problem 7 Give the re bounded between the grph of f(x) = x 2 2x nd the xxis on [ 1, 3]. Evlute 3 1 ( x 2 2x ) dx nd interpret the result in terms of res. Indefinite Integrl s Generl Antiderivtive The Indefinite Integrl of f In generl, f(x) dx = F (x) + C where F (x) is ny ntiderivtive of f(x) nd C is n rbitrry constnt. 8
9 Problem 8 1. Find F given tht F (x) = cos 3x nd F ( π) = Give n ntiderivtive of f(x) = cos 3x whose grph hs yintercept 3. Undoing the Chin Rule: The usubstitution Note: Most differentition involves the chin rule, so we should expect tht most ntidifferentition will involve undoing the chin rule [1ex] (the usubstitution) If F is n ntiderivtive for f, then d [ ] F (u(x)) = F (u(x)) u (x) = f(u(x)) u (x) dx f(u(x)) u (x) dx = f(u) du = F (u) + C = F (u(x)) + C. The usubstitution 9
10 Problem 9 Clculte 1. sin x cos x dx 2. 2x 3 sec 2 (x 4 + 1) dx 3. sec 3 x tn x dx 4. x(x 3) 5 dx Substitution in Definite Integrls The Chnge of Vribles Formul f(u(x))u (x) dx = u(b) u() f(u) du. We chnge the limits of integrtion to reflect the substitution. Problem 10 10
11 Evlute / (x 2 1)(x 3 3x + 2) 3 dx cos 3 πx sin πx dx x 5 x dx Definite Integrl nd Antiderivtive Problem Find f(x) such tht x 2 ( d x ) f(t) dt = f(x). dx f(t) dt = cos(2x) Give the function f(x) tht solves the eqution x 2 (t + 1)f(t) dt = sin(x). Properties ( d u ) f(t) dt = f(u) du dx dx ( ) d b f(t) dt = f(v) dv dx v dx ( d u ) f(t) dt = f(u) du dv f(v) dx v dx dx 11
12 Problem 12 Find ( ) d x 3 dt dx t d dx ( x 2 ( 3t sin(t 2 ) ) ) dt 3 d ( 2x dt ) dx x 1 + t 2 MenVlue Theorems for Integrls Let f vg denote the verge or men vlue of f on [, b]. Then f vg = 1 b f(x) dx. The First MenVlue Theorems for Integrls If f is continous on [, b], then there is t lest one number c in (, b) for which f(c) = f vg. Problem 13 Give the verge vlue of the function f(x) = sin x on the intervl [0, π/2]. Problem 14 Give the vlue of c tht stisfies the conclusion of the men vlue theorem for integrls for the function f(x) = x 2 2x + 3 on the intervl [1, 4]. 12
13 Are by Integrtion with Respect to x: f(x) g(x) Rectngle Are [f(x i ) g(x i )] x i Riemnn Sum [f(x i ) g(x i )] x i 13
14 re(ω) = [f(x) g(x)] dx = lim [f(x i ) g(x i )] x i. P 0 Are by Integrtion with Respect to y: F (y) G(y) Rectngle Are [F (y i ) G(y i )] y i 14
15 Riemnn Sum [F (y i ) G(y i )] y i 15
16 re(ω) = d c [F (y) G(y)] dy = lim [F (y i ) G(yi )] y i. P 0 Problem 15 Give formul involving integrl(s) in x for the region bounded by y = x 2 nd y = x. 16
17 Problem 16 Give formul involving integrl(s) in y for the region bounded by y = x 2 nd y = x. Solid of Revolution About the xaxis: Wsher Cylinder Volume: π([f(x i )]2 [g(x i )]2 ) x i [1ex] Riemnn Sum: π([f(x i )] 2 [g(x i )]2 ) x i [1ex] 17
18 V = π([f(x)] 2 [g(x)] 2 ) dx = lim π([f(x i )] 2 [g(x i )] 2 ) x i P 0 Solid of Revolution About the xaxis: Shell V = d c 2π y [F (y) G(y)] dy = lim 2π y i [F (yi ) G(yi )] y i. P 0 The integrnd 2π y [F (y) G(y)] is the lterl re of the cylinder. Problem 17 The region bounded by y = x 2 nd y = 2x is rotted round the xxis. Give formul involving integrls in x for the volume of the solid tht is generted. 18
19 Problem 18 The region bounded by y = x 2 nd y = 2x is rotted round the xxis. Give formul involving integrls in y for the volume of the solid tht is generted. Solid of Revolution About the yaxis: Wsher Cylinder Volume: π([f (y i )]2 [G(y i )]2 ) y i [1ex] Riemnn Sum: π([f (y i )] 2 [G(y i )]2 ) y i [1ex] 19
20 V = d c π([f (y)] 2 [G(y)] 2 ) dy = lim π([f (y i )] 2 [G(yi )] 2 ) y i P 0 Solid of Revolution About the yaxis: Shell V = 2π x [f(x) g(x)] dx = lim 2π x i [f(x i ) g(x i )] x i. P 0 The integrnd 2π x [f(x) g(x)] is the lterl re of the cylinder. Problem 19 The region bounded by y = x 2 nd y = 2x is rotted round the yxis. Give formul involving integrls in y for the volume of the solid tht is generted. 20
21 Problem 20 The region bounded by y = x 2 nd y = 2x is rotted round the yxis. Give formul involving integrls in x for the volume of the solid tht is generted. 21
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