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Verified answer
Integral Tentu∫3(x + 1)(x - 6) dx [2_0]
= 3 ∫(x² - 5x - 6) dx
= 3 (x³/3 - 5x²/2 - 6x)
= 3 [(2³/3 - 5 . 2²/2 - 6 . 2) - 0)]
= 3 (8/3 - 10 - 12)
= 3 (8/3 - 22)
= 8 - 66
= -58