定积分:过原点的直线l与抛物线y=x2-4x所围成的图形的面积是36,求l的方程
人气:255 ℃ 时间:2019-08-20 05:51:27
解答
设y=kx+b.
让∫(kx+b-(x^2-4x))dx=36,其中积分区域为x0,x1,x0
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