Question

微积分，求解~~~

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Answer 1

y'=-2x+4 y'(0)=4 y'(3)=-2
所以AC:y+3=4x BC:y=-2(x-3)
-2x+6+3=4x x=3/2 y=-3+6=3
所以C(3/2,3)
设BC与y轴交于M 过B垂直x轴的直线与AC交于N
M(0,6) N(3,9)
区域MABCM面积为S1=∫(0,3) -2x+6 +x^2-4x+3 dx=∫(0,3) x^2-6x+9dx=(x^3/3 -3x^2+9x)/(0,3)=9
区域ACNBA面积为S2=∫(0,3) 4x-3+x^2-4x+3dx=∫(0,3) x^2dx=(x^3/3)/(0,3)=9
SACM=1/2*3/2 *(3+6)=27/4 S BCN=1/2*(3-3/2)*9=27/4
所以所求的区域面积S =1/2(S1+S2-SACM-SBCN)=1/2(9+9-27/4-27/4)=1/2(18-27/2)=9/4

Answer 2

解
y=-x^2+4x-3,y'=-2x+4,y'(0)=4.y'(3)=-2
A处切线 y+3=4x(1)
B处切线 y=-2(x-3)(2)
（1），（2）联立x=3/2,y=3即C(3/2，3）
所求面积=S_0^(3/2)(4x-3+x^2-4x+3)dx+S_{3/2}^3(-2x+6+x^2-4x+3)dx
=S_0^(3/2)x^2dx+S_{3/2}^3(x^2-6x+9)dx
=1/3(3/2)^3+[1/3(27-(3/2)^3)-3(9-(3/2)^2)+9*3/2]
=9/8+[189/24-81/4+27/2]
=[27+189-486+324]/24=54/24=9/4

Answer 3

2.25 .