※ 引述《tiffany9802 (Tiffany)》之銘言：
: 如題
: 1
: S xe^xdx?
: -1
: 請問一下答案是e^1-e^-1嗎??
1
∫ (x)(e^x) dx
-1
|1 1
= (x)(e^x) | - ∫ e^x dx
|-1 -1
(令 u = e^x , dv = dx , 則 du = e^x dx , v = x)
|1
= (1)(e^1) - (-1)(e^(-1)) - (e^x) |
|-1
= e + e^(-1) - (e - e^(-1))
2
= e + e^(-1) - e + e^(-1) = (2)(e^(-1)) = ---
e
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