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Ksivusya [100]
3 years ago
13

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

Mathematics
1 answer:
REY [17]3 years ago
8 0
<span>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</span>
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