Question

求由曲线Y=X²与直线Y=X+2所围成的平面图形的面积

Answer 1

如图所示

Answer 2

y=x^2
y=x+2
x^2=x+2
x^2-x-2=0
(x-2)(x+1)=0
x=-1 or 2
A
=∫(-1->2) (x+2 -x^2) dx
=[(1/2)x^2+2x-(1/3)x^3]|(-1->2)
=(2 +4 - 8/3) -( 1/2 -2 +1/3)
=6 +2 -8/3 -5/6
=8 - 7/2
=9/2