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w5squarerootw31
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K
integral (w^5)(squareroot((w^3)-1)
u= w^3 du= 3w^2dw integral 1/3[u (u-1)^1/2 du 1/3 [ u^3/2 -udu] 1/3 [[ 2/5(u^5/2) - 1/2u^2]] the answer is 1/3[2/5(u^5/2) - 2/3(u^3/2) what did I do wrong in the second part?
khuezy
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Jun 7, 2008
Replies: 2
Forum:
Calculus
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