若∫12f(x)dx=1\int_{1}^{2}f(x)dx = 1∫12f(x)dx=1,∫12g(x)dx=2\int_{1}^{2}g(x)dx = 2∫12g(x)dx=2,则∫12[f(x)+2g(x)]dx=\int_{1}^{2}[f(x)+2g(x)]dx =∫12[f(x)+2g(x)]dx=( ).
(A)333
(B)444
(C)555
(D)666
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