Re: [微積] 台大研究所微積分考試
-1 ┌ 2 y ┐
Let y = g(x), then f(y) = g (y) = x and │(2y + y ) e │dy = f'(y) dy = dx.
└ ┘
y = 0 => f(0) = 0; y = 1 => f(1) = e.
Thus,
╭ e ╭ 1 ┌ 2 y ┐ 3 2 y │y=1
│ g(x) dx = │ y│(2y + y ) e │dy = (y - y + 2y - 2) e │ = 2.
╯0 ╯0 └ ┘ │y=0
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