求由抛物线y=4-x^2 ,及在点(2,0) 处的切线和y 轴所围成的平面图形的面积

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y' = -2x
x = 2,y' = -4
切线:y - 0 = -4(x - 2),y = 8 - 4x
x = 0,y = 8,切线与轴的交点为(0,8)
S = ∫₀²(8 - 4x - 4 + x²)dx
= (4x - 2x² + x³/3)|₀²
= 8/3