4.3 - MA 160 ws 4.3 Area and Definite Integrals Name G o r...

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Unformatted text preview: MA 160 ws 4.3 Area and Definite Integrals Name G o, r G u z to n Definition of definite integral. Let f be any continuous function over the interval [a, b] and F any antlderlvatlve of f. Then the definite Integral b offfromatobis If(x)dx=F(b)—F(a) . Fundamental Theorem of Integral Calculus: n [a, b}, then lim 2 f(x,-)Ax =J‘: f(x)dx = F(x)|: = F(b)— F(a). n—No . 1:1 H a continuous function f has an antiderivative F over 1. Set up the integral that gives the area of the shaded region. The equation of the curve is y = if; +1 . 3 3 3 J... 3‘ -L,X -l 3 _1__ ‘3 j(9x+‘)‘1"'q 3'1“" 'K(3)+3-(”_(-15~+{- : 531+: +31%? 2.31%., =35'*60..Z§ l (.2 [A I; ‘1’; 2. Shade the region whose area is given by [14(4xi x2) (it. ”waffle. (3 Gd d' [13036-32 dx :f’aflxi I2 x + ‘iiel x 3 1 3 3 a. l a : “(:5 ._é(3rm(g)u(g—-—¢+cz) Continued9 : 3&‘5’9f27-«é—1—é-9 1 2 (2-1}:194: _ (q 3 3 4. Find the area under each of the given curves over the given interval: a. _ = J— ,4,9 b. =x2 , —2,1] A? v x x9 I” ] A; ll 3, .L I3 4):): 1‘le jqxquxiszdx 5.1.x AX:‘:€'X| 3( "( 3 I *2. ' _.;2x57’-q_ at“ 5:; 51 -1376” | ”H'— _H(q -LI 2.) 2: I 5 ‘f 51 ; )_IiE-S_l _ “§(Q‘I3*32)3'5'(2” * 5 c y 3 11,61 4 d. y=e°5x ,HAI 4, .5x .5'x Ll- 5%dx:3&x{é ite dx::}§-€ I f l 'l =3(L-te—,enll : 52(e'6YQL .51.”) Z a “‘2. 3‘0“; :3(e“6‘)=9(ea-J:) V6 e. y=2x+—xl_7 ,[l,4] )L‘L Q jq(Qx+-'——)dx : 597‘; 1.25;! :: X31— —'—— z: Lt: J.»_(II__ x- r 3) , x p. a; i x q ' -lé—~4-o;}5 4 I .. @3 LI. 5. A company estimates that its sales will grow continuously at a rate given by the function S'(t) z 209% , where S'(t) is the rate at which sales are increasing, in dollars per day, on day t. a. Find the accumulated sales for the first five days. 5' 5:8:(£)th asaoeaifidt : ‘3 0' 7:1 84:19 :po0 (6 57: Ca") : ‘40 (ea :1) " 3‘ 997. 30 b. Find the sales from the 2"Id day through the 5th day. (That is integrate from 1 to 5.) foam— : 315.206“; '3 1.: Q0 _ i- e-‘ae ’3; go (a 5;: etc) 1- I 6. Find s(t) ifa(t) = *3 + 6,with v03) = 6 and 3(0) = 10. Vlflfilaéfl‘Lf: £021"chth = ‘“ 32: éffl-C, I z: -t iota—Cl LL44. w ' '1 $06) a) :é A” 52: *oz-réCa‘l*C—i ”CrréMvCélz—f"+et+6. =fvétldt = IGt‘fie-eeMDJ at = ——;—f3+e.§‘+ée+ca=-§e3+ SfZ-E-é-éi-C. ML .SCcflrrofigd ja:-——f 3 ‘3‘ , >. a “+50 +é~o+CA-4&C;:;o_ 2 W: $g£):_:;_t3+3tat+é&+’o‘ Answers for WS 4.3: 3 2 1. (fi-+l)dfx 2. —2 e4—1 21.5. 15. 3a. 27 3b. 4 3c. 2 3d. 3 4a. 5135-2- 4b. 3 4c.3ln6 4d. 2[el——1—]4e. 9-3- JE 4 5a. $447.30 5b. $421.35 6.3(t)=—lt3+3r2+61+10 3 ...
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