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28 March, 07:30

Evaluate the integral. 2 (6t i - t3 j + 2t3 k) dt 0

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  1. 28 March, 08:18
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    2 (6ti-t3j+2t3k) dt 0 = 2 (6 (1/2) t2i-1/4t4j+2 (1/4) t4k) 0 = 6 (1/2) (0) 2i-1/4 (0) 4j+2 (1/4) (0) 4k-6 (1/2) (2) 2i-1/4 (2) 4j+2 (1/4) (2) 4k = - 12i+4j-8k
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