计算函数y=4-x²和y=x（x-2）所围成图形面积？ 5
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y=4-x^2 (1)
y=x(x-2) (2)
from (1) and (2)
4-x^2 =x(x-2)
2x^2-2x-4=0
x^2-x-2=0
(x+1)(x-2)=0
x=-1 or 2
A
=∫(-1->2) [ (4-x^2) -x(x-2) ] dx
=∫(-1->2) (4+2x -2x^2) dx
=[4x+x^2 - (2/3)x^3]|(-1->2)
=(8+4-16/3) -(-4+1+2/3)
=12-16/3 +3 -2/3
=15-6
=9
y=x(x-2) (2)
from (1) and (2)
4-x^2 =x(x-2)
2x^2-2x-4=0
x^2-x-2=0
(x+1)(x-2)=0
x=-1 or 2
A
=∫(-1->2) [ (4-x^2) -x(x-2) ] dx
=∫(-1->2) (4+2x -2x^2) dx
=[4x+x^2 - (2/3)x^3]|(-1->2)
=(8+4-16/3) -(-4+1+2/3)
=12-16/3 +3 -2/3
=15-6
=9
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