APPLICATIONS OF THE DEFINITE INTEGRAL 23 Here f x 10 e x 20 e x 20 f x 10 20 e

Applications of the definite integral 23 here f x 10

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APPLICATIONS OF THE DEFINITE INTEGRAL23.Heref(x) = 10ex/20+e-x/20f0(x) =1020ex/20-e-x/201 +f0(x)2= 1 +12ex/20-e-x/202=12ex/20+e-x/202Now,s=20Z-2012ex/20+e-x/20dx=20Z0ex/20+e-x/20dx= 20ex/20-e-x/20200= 20(e-e-1)47.008024.s=Z30-30s1 +12(ex/30-e-x/30)2dx=Z30-3012ex/30+e-x/30dx=15ex/30-15e-x/3030-30= 30e-30e-170.51207161ft.25.In Example 4.4,y(x) = 5(ex/10+e-x/10)y(0) = 5(e0+e0) = 10y(-10) =y(10)= 5(e1+e-1) = 15.43sag = 15.43-10 = 5.43 ftA lower estimate for the arc length given thesag would be2p(10)2+ (sag)2= 2100 + 29.484922.76This looks good against the calculated arclength of 23.504.26.Ifx2/3+y2/3= 1, then in the first quad-rant,y= (1-x2/3)3/2and taking only thefirst-quadrant case (which would produce onefourth of the total lengths), we havey=32(1-x2/3)1/2-23x-1/3=-x-1/3(1-x2/3)1/2(y0)2=x-2/3(1-x2/3) =x-2/3-1s= 4Z10p1 +y02dx= 4Z10px-2/3dx= 4Z10x-1/3dx= 432x2/310= 6There are some technicalities in fully justifyingthe preceding computation, since the integrand(x-1/3) is unbounded atx= 0, but the con-clusion is sound.xy10127.y= 0 whenx= 0 and whenx= 60, so thepunt traveled 60 yards horizontally.y0(x) = 4-215x=215(30-x)This is zero only whenx= 30, at which pointthe punt was (30)2/15 = 60 yards high.s=Z600s1 +4-215x2dx139.4 yardsv=s4 sec=139.4 yards4 sec·3 feet1 yard= 104.55 ft/s604005030x3050102004020601028.Sincey(100)=0,theballtraveled100yards.The maximum height of the ball isy(50) =253yards.The arc length iss=Z1000s1 +1300(100-2x)2dx101.82215 yards
5.4.ARC LENGTH AND SURFACE AREA33184620x40600802010029.S= 2πZ10y ds= 2πZ10x2p1 + (2x)2dx3.809730.S=Zπ02πsinxp1 + cos2xdx14.4236031.S= 2πZ20y ds= 2πZ20(2x-x2)p1 + (2-2x)2dx10.965432.S=Z0-22π(x3-4x)p1 + (3x2-4)2dx67.0655733.S= 2πZ10y ds= 2πZ10exp1 +e2xdx22.943034.S=Z212πlnxr1 +1x2dx2.8656335.S= 2πZπ/20y ds= 2πZπ/20cosxp1 + sin2xdx7.211736.S=Z212πxr1 +14xdx8.2831537.s1=Z10q1 + (6x5)2dx=Z10p1 + 36x10dx1.672s2=Z10q1 + (8x7)2dx=Z10p1 + 64x14dx1.720s3=Z10q1 + (10x9)2dx=Z10p1 + 100x18dx1.75Asn→ ∞, the length approaches 2, since onecan see that the graph ofy=xnon [0,1] ap-proaches a path consisting of the horizontalline segment from (0,0) to (1,0) followed bythe vertical line segment from (1,0) to (1,1).38.(a) For 0x <1, we havelimn→∞xn= 0Therefore, the length of the limiting curveis 1 (the limiting curve is a horizontalline).Connecting the limiting curve tothe endpoint at (1,1) adds an additionallength of 1 for a total length of 2.(b)y1=x4, y01= 4x3y2=x2, y02= 2xSince both are increasing for positivex,y1is “steeper” (y2is “flatter”) if and only ify01> y02, i.e.,4x3>2x, x2>12, x >r1239.(a)L1=Zπ/6-π/6p1 + cos2xdx1.44829L2=ssinπ6-sin-π62+2π621.44797 HenceL2L1=1.447971.44829.9998(b)L1=Zπ/2-π/2p1 + cos2xdx3.8202L2=r2 sinπ22+ (π)2=pπ2+ 4 = 3.7242HenceL2L10.974940.(a)L1=Z53p1 + (ex)2dx128.3491L2=p22+ (e5-e3)2128.3432HenceL2L10.9999
332CHAPTER 5.APPLICATIONS OF THE DEFINITE INTEGRAL(b)L1=Z-3-5p

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