y=-x^2+2与直线y=x交点为:
x=-x^2+2
x^2+x-2=0
(x+2)(x-1)=0
x=-2或x=1
即交点为(-2,-2)(1,1)
所以
面积=∫(-2,1)(-x^2+2-x)dx
=(-x^3/3+2x-x^2/2)|(-2,1)
=(-1/3+2-1/2)-(8/3-4-2)
=7/6+6-16/6
=6-9/6
=6-3/2
=9/2
选D
y=-x^2+2与直线y=x交点为:
x=-x^2+2
x^2+x-2=0
(x+2)(x-1)=0
x=-2或x=1
即交点为(-2,-2)(1,1)
所以
面积=∫(-2,1)(-x^2+2-x)dx
=(-x^3/3+2x-x^2/2)|(-2,1)
=(-1/3+2-1/2)-(8/3-4-2)
=7/6+6-16/6
=6-9/6
=6-3/2
=9/2
选D