Dr. Z’s Math152 Handout #6.1 [Areas Between Curves]By Doron ZeilbergerProblem Type 6.1a : Find the area of the region enclosed between y = Expr1(x) and y =Expr2(x), where they meet at two points.Example Problem 6.1a:Find the area of the region enclosed between y = 12 − x2and y = x2− 6.Steps Example1. Find the points of intersection bysolving, for x the algebraic equationExpr1(x) = Expr2(x) .If you get just one root, then the region isinfinite, and there is no (finite) answer. Ifyou get two roots, let’s call them x1< x2,then the limits of the integral will be x1and x2. If you get three roots, see Prob-lem 6.1b below.1. 12 − x2= x2− 6 means 2x2= 18, i.e.2(x2− 9) = 0 so x2− 9 = 0, which means(x − 3)(x + 3) = 0, and we have two rootsx = −3 and x = 3.2. For each of the intervals of integration(in this example there is just one), deter-mine which curve in on top, by plugging-in a random value (or sketching).2. Plugging-in x = 0 into y = 12 − x2gives y = 12, while plugging-in x = 0 intoy = x2−6 gives y = −6. Since 12 is biggerthan −6, we have that T OP = 12−x2andBOT = x2− 6.3. The area is the integralZx2x1[T OP (x) − BOT (x)] .3. The area isZ3−3[(12−x2)−(x2−6))] =Z3−3[18−2x2] =18x−2x33|3x=−3= 18(3−(−3))−2(33− (−3)3)3= 108−36 = 72 .Ans.: 72.1Problem Type 6.1b : Find the area of the region enclosed between y = Expr1(x) and y =Expr2(x), where they meet at three points.Example Problem 6.1b:Find the area of the region enclosed between y = x3− x and y = 3x.Steps Example1. Find the points of intersection bysolving, for x the algebraic equationExpr1(x) = Expr2(x) .If you get one root, then the region is in-finite, and there is no (finite) answer. Ifyou get two roots, let’s call them x1< x2,then the limits of the integral will be x1and x2(like in 6.1a). If you get threeroots, x1< x2< x3, then you need todo TWO integrals, oneRx2x1and the otherRx3x2.1. x3− x = 3x means x3− 4x = 0, i.e.x(x − 2)(x + 2) = 0 so we have three rootsx = −2, x = 0 and x = 2.2. For each of the intervals of integration(in this example there are two: (x1, x2)and (x2, x3)), determine which curve is ontop, by plugging-in a random value (orsketching).2. Plugging-in x = −1 into y = x3− xgives y = 0, while plugging-in x = −1into y = 3x gives y = −3. Since 0 is big-ger than −3, we have that on the interval(−2, 0), T OP = x3− x and BOT = 3x.Regarding the interval (0, 2), plugging-inx = 1 into y = x3− x gives y = 0, whileplugging-in x = 1 into y = 3x gives y = 3.Since 3 is bigger than 0, we have that onthe interval (0, 2), TOP = 3x and BOT =x3− x.23. The area is the sum of the integralsZx2x1[T OP (x) − BOT (x)]+Zx3x2[T OP (x) − BOT (x)] ,note that who is TOP and who is BOT(usually) gets switched between these in-tervals.3. The area isZ0−2[(x3−x)−(3x)] +Z20[(3x)−(x3−x)] =Z0−2[x3− 4x] +Z20[−(x3− 4x)] =[x4/4−2x2]|0−2+[−(x4/4−2x2)]|20= 4+4 = 8 .Ans.: