Question: Solution by Method I: (2x - 1)3dx; let u = 2x -1; du = 2dx or -du = dx (2x - 1)3 dx = u3

Solution by Method I: (2x - 1)3dx; let u = 2x -1; du = 2dx or -du = dx (2x - 1)3 dx = u3 (zdu) 1 ub 7 4 a 7 (2x - 1)+ 12 = [(2 - 2- 1)*] -=[(2 - 1-1)4 ] = -(3) (1) 4 = (34 - 1)+ = 10By Method II: cosBy sin ydyu = cos y TE / 3 du = -sin ydy -du = sin ydy when y = 3 ' 1 = COS when y = 7 1 = COS - = 0 TE/ 2 cos By sin ydy = u3 (-du) IT / 3 1/2 0 = -4 1/2 = 64

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