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$\int_{0}^{6} x (x - 1) (x - 2) (x - 3) (x - 4) (x - 5) (x - 6) d x$

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Correct option is A

The point of symmetry is (3,0) We make the change of variable x=t+3 Then g(t)=(t+3)(t+2)(t+1)t(t−1)(t−2)(t−3)∴∫06 x(x−1)(x−2)(x−3)(x−4)(x−5)(x−6)dx=∫−33 g(t)dt but g(−t)=−g(t) therefore g(t) is odd function =0

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∫06xx−1x−2x−3x−4x−5x−6dx

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$\int_{0}^{6} x (x - 1) (x - 2) (x - 3) (x - 4) (x - 5) (x - 6) d x$