Question

求解题过程～ 高数定积分应用

Answer 1

1.    联立解 y = x^2+1,  y = 2x+3,  得交点P(1-√3, 5-2√3),   Q(1+√3, 5+2√3)

   S = ∫<1-√3, 1+√3> [(2x+3)-(x^2+1)]dx = ∫<1-√3, 1+√3> (2+2x-x^2)dx

   = [2x+x^2-x^3/3]<1-√3, 1+√3> = 4√3

   V = π∫<1-√3, 1+√3> [(2x+3)^2-(x^2+1)^2]dx

   = π∫<1-√3, 1+√3> (8+12x+2x^2-x^4)dx

   = π[8x+6x^2+(2/3)x^3-x^5/5]<1-√3, 1+√3>(152/5)π√3

2. 另题类似可仿算。