Question

计算 y=3-2x-x^2 及 y=0 所围成的面积？

Answer 1

y=3-2x-x^2=-(x+3)(x-1) 交点是(1,0),(-3,0)
∫（-3，1）(3-2x-x^2)dx=(-1/3*x^3-x^2+3x)|（-3，1）=-1/3-1+3-(9-9-9)=11-1/3=32/3 望采纳

Answer 2

y=3-2x-x^2
y=0
3-2x-x^2 =0
x^2+2x-3=0
(x+3)(x-1)=0
x=-3 or 1
A
=∫(-3->1) (3-2x-x^2) dx
=[3x-x^2-(1/3)x^3]|(-3->1)
=(3-1-1/3)-(-9-9+9)
=2-1/3 +9
=32/3