hw-4-sec-6.1-solns

# hw-4-sec-6.1-solns - 3.A 6 El APPLICATIONS OF INTEGRATION...

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Unformatted text preview: 3.A 6 El APPLICATIONS OF INTEGRATION 6.1 Areas Between Curves I! ll H H The curves intersect when ac : a: z(m—1)=O ¢> :L‘zﬂorl. \$=4 ll (yr—y3)d1:=/D [(Sm—m2)—x]dx=/O(4m—\$2)dx [2:02— 1:3] =(32—63—4)——(0)=§3Z 3‘53] =108—72236 y=1 -y2+1)dy=[ y3/2—%y3+y 0 (\$R'IL)dy:A1[ﬁ_(yz +1)—(0) 03h“ <§~ 3 [(2.21 — 2/2) ~ (y? ~ 4w} dy = /O (-2212 + 6y) dy : [—ézﬁ + 3.1/2? = (—18 + 27) — 0 = 9 _§2 Ir:2(:cwsinx)dx = [ H —- + c051? w/2 —-——1 ZL+0 c» 32—2: duh—- \$664 A ecurvesintersectwhenri’—2m=:c+4 <27 x2—32—420 <3 (I+1)(CL‘—4)=O 42> xz—lor4. 426 U CHAPTERS APPLICATIONS OF INTEGRATION 4 A=/ [1+4—(m2—~2x)jdr —1 4 =/ (—x2+3x+4)dx —1 [*W + 3:62 +455]: H " ’1‘ «42 + t0 p + H 3} l A (uh-d 4.. wlw I i: H 9. First ﬁnd the points of intersection: V1 + 3 z 1‘ T 3 :> (V’x + 3)2 = < 2 2 > => z+3=§(x+3)2 => 2 4(x+3)—(m+3)220 :> (r+3){4—(5E+3)]=0 3 (m+3)(1—\$):0 =¢ x=—3or1.So [42/53 (m—x+3)dx 2 : [.3437 + 3)3/2 - 1 4 -3 1o.1+\/‘=3:1=1+ Dela 1, 11. A nguwm. u—a-nngén 428 D CHAPTERS APPLICATIONS OF INTEGRAT|ON 16.x3—x=3w <=> \$3-4m=0 4=> x(x2—4)=0 «t? a:(a:+2)(:c——2)=O 4:) a: = 0, —2, or 2. A:/2 I3x—(w3—m)ld:c 2 = 2 A [33; -— (m3 —- ﬂ] d2: [by symmetry] = 2/2012 —a:3)d\$ = 2[2a:2 — i334]: = 2(8 -4) z 8 o é-rczﬁ => im2=x => z2—4x=0 => a:(m—4)=O A=/4(\/§—lm)dx+/g (lx~\/§)dr=[3m3ﬂ—i\$2] +[lm — 2 2 3 4 4 a 0 4 0 =[(%—4)—0]+[(§}“18)*(4‘?)]=7+3 Sac" We start by ﬁnding the equation of the tangent line to y = 2:2 at the int 1, 1 : p0 436 D CHAPTERS APPLICATIONS OF INTEGRATION y’ = 2x, so the slope of the tangent is 2(1) = 2, and its equation is y - 1 = 2(ac —— 1), or y = 2:1: —— 1. We would need two integrals to integrate with respect to x, but only one to integrate with respect to y. 1 A: fo1 am 1) — ﬂ] dy: [gyz + %y_ gym] +£— 0 .1. 12 who 1 4 By the symmetry of the problem, we consider only the ﬁrst quadrant, where y = 3:2 :> 1’ 2 V3 We are looking for a number b such that fobﬁdy=fb4ﬂdy => §[y3/2]b:§[y3/2]4 :> 0 b 123/2 = 43/2 ~ b3” :> 21;”2 = 8 => (23/2 = 4 => (2 = 42/3 z 2.52. _1" .14 1 1 1 5 —- = —— :r ——+1=——+- => -= 2: :5 a 4 a 4 ll => 91w 9 II 01100 (b) The area under the curve y = 1/322 from m : 1 to a: = 4 is % [take a = 4 in the ﬁrst integral in part (3)]. Now the line y = I) must intersect the curve a: = 1/ ﬂ and not the line a: = 4, since the area under the line 3; = 1 //-12 from a; = 1 to as = 4 is only %, which is less than half of § . We want to choose b so that the upper area in the diagram is half of the total area under the curve y 2 1/1132 from 1‘ = 1 to a: = 4. This implies that fill/ﬂ-1)dy=%‘% => [2x/ﬁ~yli=% => 1e2¢5+b=2 => bat/hi =0. Lettirigc=\/E,wegetc2—2c+§ :0 => 8 8 862—16c+5=0.Thus,c=3§£Q13g£-::@:1i-@.Butc=\/5<1 => c=1—>§ => b:c2=1+%——‘é§=§(11~—4\/§)z0.1503i 51. We ﬁrst assume that c > 0, since c can be replaced by —c in both equations without changing the graphs, and if c = O the curves do not enclose a region. We see i'rom‘the graph that the enclosed area A lies between ac = —c and a: = c, and by symmetry, it is equal to four times the area in the ﬁrst quadrant. The enclosed area is A = 4f: (62 — x2)da; = 4[c2:c — \$53]: = 4(c3 — §c3) = 4(%c3) = §c3 SoA=576 <=> gca =576 <=> c3 2:216 ¢:> c: €/216=6. Note that c = ——6 is another solution, since the graphs are the same. ...
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## This note was uploaded on 10/10/2011 for the course M 408S taught by Professor Shirley during the Spring '11 term at UT Arlington.

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hw-4-sec-6.1-solns - 3.A 6 El APPLICATIONS OF INTEGRATION...

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