求由曲线y=x^2,y=x+1,所围平面图形的面积 答案是4.5。求大神指导

Question

如题所述

Answer 1

解：两曲线交点：A[(1+√5)/2 ，(3+√5)/2] ；B[(1-√5)/2 ， (3-√5)/2]
ds=[(x+1)-x²]*dx
S=∫ds《xb∽xa》=∫[(x+1)-x²]dx|(for xb to xa)
=-x³/3+x²/2+x|(1-√5)/2 ， (1+√5)/2
=[-(1+√5)³/24+(1+√5)²/8+(1+√5)]-[-(1-√5)³/24+(1-√5)²/8+(1-√5)]
=(-1/24)*[(1+√5)³-（1-√5)³]+(1/8)*[(1+√5)²-(1-√5)²]+[(1+√5)-(1-√5)]
=-2√5/3+√5/2+2√5=11√5/6 【≈4.099】