Question

18．求${∫}_{-1}^{1}$f（x）dx，其中f（x）=$\left\{\begin{array}{l}{2x-1，-1≤x＜0}\\{{e}^{-x}，0≤x≤1}\end{array}\right.$且${∫}_{-1}^{0}$（2x-1）dx=-2，${∫}_{0}^{1}$e^-xdx=1-e^-1．

Answer 1

分析 由${∫}_{-1}^{1}$f（x）dx=${∫}_{-1}^{0}$（2x-1）dx+${∫}_{0}^{1}$e^-xdx，代值計算即可．

解答 解：∵f（x）=$\left\{\begin{array}{l}{2x-1，-1≤x＜0}\\{{e}^{-x}，0≤x≤1}\end{array}\right.$且${∫}_{-1}^{0}$（2x-1）dx=-2，${∫}_{0}^{1}$e^-xdx=1-e^-1，
∴${∫}_{-1}^{1}$f（x）dx=${∫}_{-1}^{0}$（2x-1）dx+${∫}_{0}^{1}$e^-xdx=-2+1-e^-1=-1-$\frac{1}{e}$．

點評 本題考查分段函數(shù)的定積分的問題，關(guān)鍵時掌握運算法則，屬于基礎(chǔ)題．