Re: [微積] 積分問題
u = x - 1
f(x) = x^2
f(u) = (u + 1)^2
1
∫2πuf(u)du
0
= 2π∫u(u^2 + 2u + 1)du
1
= 2π[(1/4)u^4 + (2/3)u^3 + (1/2)u^2]|
0
= 2π[(1/4) + (2/3) + (1/2)]
= (17/6)π
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