5612
6512
12
65
Correct answer is B
1∫0x2(x3+2)3dx
let u=x3+2,du=3x2dx
when x = 1, u = 3
when x = 0, u = 2
dx = du3x2
3∫2 x2[u]33x2
3∫2 u33 du
= u43∗423
112[u4]23
112[34−24]
112[81−16]
6512
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