# Determine the area of a 1 area a 13 2 4 ln 4 2 area a

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Determine the area ofA.1.area(A) =132-4 ln 42.area(A) =172-4 ln 43.area(A) =172+ 4 ln 44.area(A) =152-4 ln 4correct5.area(A) =132+ 4 ln 46.area(A) =152+ 4 ln 4Explanation:If the graph ofx(t) =t-1t,y(t) = 2t+2tintersectsy= 5 whent=t0andt=t1, thenAis similar to the shaded region in5xyabwitha=x(t0) andb=x(t1). In this casearea(A) =integraldisplayba(5-y)dx=integraldisplayx(t1)x(t0)parenleftBig5-2t-2tparenrightBigdx ,wherex=x(t) =t-1t,dx=parenleftBig1 +1t2parenrightBigdt .Thus after a change of variable fromxtotwereduce the integral describing area(A) tointegraldisplayt1t0parenleftBig5-2t-2tparenrightBigparenleftBig1 +1t2parenrightBigdt=integraldisplayt1t0parenleftBig5-2t-4t+5t2-2t3parenrightBigdt=bracketleftBig5t-t2-4 lnt-5t+1t2bracketrightBigt1t0.
mehmood (ajm4462) – Homework 10.2 – karakurt – (56295)6But the graph ofx(t) =t-1t,y(t) = 2t+2tintersectsy= 5 when2t+2t= 5,i.e., when2t2-5t+ 2 = (2t-1)(t-2) = 0.Thust0= 1/2 whilet1= 2, soarea(A) =parenleftBig154-4 ln 2parenrightBig-parenleftBig-154-4 lnparenleftBig12parenrightBigparenrightBig.Consequently,area(A) =152-4 ln 4.keywords: parametric curve, area,01110.0pointsWhich one of the following integrals givesthe length of the parametric curvex(t) =t2,y(t) = 2t ,0t8.1.I=integraldisplay80|t2+ 1|dt2.I= 2integraldisplay40|t2+ 1|dt3.I=integraldisplay40|t2+ 1|dt4.I= 2integraldisplay80radicalbigt2+ 1dtcorrect5.I=integraldisplay80radicalbigt2+ 1dt6.I= 2integraldisplay40radicalbigt2+ 1dtExplanation:The arc length of the parametric curve(x(t), y(t)),atbis given by the integralI=integraldisplaybaradicalBig(x(t))2+ (y(t))2dt .But whenx(t) =t2,y= 2t ,we see thatx(t) = 2t ,y(t) = 2.Consequently, the curve hasarc length = 2integraldisplay80radicalbigt2+ 1dt.