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求抛物線y=x²;-1與直線x=2,y=0所圍成的面積. 設題目所求面積為S,則可用定積分來求S; 函數f(x)=x²;-1的原函數為F(x)=(x³;/3)-x, 則S=∫(2,1)|(x³;/3)-x=[(2³;/3)-2]-[(1³;/3)-1]=4/3
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