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SOLUTION 29: Compute the area of the region enclosed by the graphs of the equations y=x22x+2  (or  x=1±y1) and y=2x+2  (or  x=12y1).Beginbyfindingthepointsofintersectionofthethetwographs.From y= x^2-2x+2 and y=2x+2 wegetthatx22x+2=2x+2  x24x=0  x(x4)=0    x=0  (and  y=2)  or  x=4  (and  y=10)$ Now see the given graph of the enclosed region.

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a.) Using vertical cross-sections to describe this region, we get that 0x4  and  x22x+2y2x+2 , so that the area of this region is AREA=40(Top  Bottom) dx =40((2x22x+2)(2x+2)) dx =40(2x24x) dx =(23x32x2)|40 =(23(4)32(4)2)(23(0)32(0)2) =(128332)(0) =1283963 =323 b.) Using horizontal cross-sections to describe this region, which is made up of two smaller regions, we get that 1y2  and  1y1x1+y1 in addition to 2y10  and  12y1x1+y1 , so that the area of this region is AREA=21(Right  Left) dy+102(Right  Left) dy =21((1+y1)(1y1)) dy+102((1+y1)(12y1)) dy =212y1 dy+102(2+y112y) dy =(43(y1)3/2)|21+(2y+23(y1)3/214y2)|102 =(43((2)1)3/243((1)1)3/2)+(2(10)+23((10)1)3/214(10)2)(2(2)+23((2)1)3/214(2)2) =(43(1)3/243(0)3/2)+(20+23(9)3/225)(4+23(1)3/21) =(43)+(23(27)5)(3+23) =43+13323 =23+303 =323

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