diketahui 1∫p (3t²+6t-2) dt=14 nilai-4p

InTegraL

∫(3t² + 6t - 2) dt [p...1] = 14
t³ + 3t² - 2t [p...1] = 14
p³ + 3p² - 2p - (1 + 3 - 2) = 14
p³ + 3p² - 2p = 16
p³ + 3p² - 2p - 16 = 0
(p - 2)(p² + 5p + 8) = 0
p - 2 = 0 ---> p = 2
atau
p² + 5p + 8 = 0 (akar2 imajiner)

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-4p = -4 × 2 = -8