已知函数f(x)f(x)f(x)在区间[0,2][0,2][0,2]上连续,且∫02xf(x)dx=4\int_{0}^{2}x f(x)dx = 4∫02xf(x)dx=4,则∫02f(x)dx=\int_{0}^{2}f(\sqrt{x})dx =∫02f(x)dx=( ).
(A)2
(B)4
(C)6
(D)8
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