求y=2x与y=3-x^2围成的面积既有x轴上也有x轴下.怎么理解
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y=2x (1)
y=3-x^2 (2)
(1)=(2)
2x=3-x^2
x^2+2x-3=0
(x-3)(x+1)=0
x=-1 or 3
x=-1 , y=-2
x=3, y=6
A = ∫( -2->6) (x1-x2) dy
=∫( -2->6) [ y/2 - √|3-y| ]dy
= (1/4)[y^2]| ( -2->6) -∫( -2->3) √(3-y) dy-∫(3->6) √(y-3) dy
=8 + (2/3)[ (3-y)^(3/2) ]( -2->3) - (2/3)[ (y-3)^(3/2) ](3->6)
= 8 + (2/3)(-1) - (2/3).3^(3/2)
= (22/3) - 2√3
y=3-x^2 (2)
(1)=(2)
2x=3-x^2
x^2+2x-3=0
(x-3)(x+1)=0
x=-1 or 3
x=-1 , y=-2
x=3, y=6
A = ∫( -2->6) (x1-x2) dy
=∫( -2->6) [ y/2 - √|3-y| ]dy
= (1/4)[y^2]| ( -2->6) -∫( -2->3) √(3-y) dy-∫(3->6) √(y-3) dy
=8 + (2/3)[ (3-y)^(3/2) ]( -2->3) - (2/3)[ (y-3)^(3/2) ](3->6)
= 8 + (2/3)(-1) - (2/3).3^(3/2)
= (22/3) - 2√3
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