求第五题。
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y= (1/4)x^2 +2 (1)
x-2y+8 =0 (2)
from (1) and (2)
(1/2)(x+8)=(1/4)x^2 +2
2(x+8)=x^2 +8
x^2-2x-8=0
(x-4)(x+2)=0
x=-2 or 4
A
= ∫ (-2->4) { (1/2)(x+8) - [(1/4)x^2 +2] } dx
= ∫ (-2->4) [ -(1/4)x^2 +(1/2)x +2 ] dx
= [-(1/12)x^3 + (1/4)x^2 + 2x ]|(-2->4)
=( -64/12 + 4 + 4) -(8/12 +1-4)
=5
x-2y+8 =0 (2)
from (1) and (2)
(1/2)(x+8)=(1/4)x^2 +2
2(x+8)=x^2 +8
x^2-2x-8=0
(x-4)(x+2)=0
x=-2 or 4
A
= ∫ (-2->4) { (1/2)(x+8) - [(1/4)x^2 +2] } dx
= ∫ (-2->4) [ -(1/4)x^2 +(1/2)x +2 ] dx
= [-(1/12)x^3 + (1/4)x^2 + 2x ]|(-2->4)
=( -64/12 + 4 + 4) -(8/12 +1-4)
=5
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