Question

2．若f（x）=$\left\{\begin{array}{l}{x^2}+3\\-x\end{array}\right.\begin{array}{l}x≥0\\ x＜0\end{array}$，則$\int_{-1}^1$f（x）dx=$\frac{5}{6}+\frac{2}{ln3}$．

Answer 1

分析 根據(jù)函數(shù)各段的自變量范圍將定積分表示-1到0以及0到1上的定積分的和，分別計(jì)算定積分值即可．

解答 解：f（x）=$\left\{\begin{array}{l}{x^2}+3\\-x\end{array}\right.\begin{array}{l}x≥0\\ x＜0\end{array}$，則$\int_{-1}^1$f（x）dx=${∫}_{-1}^{0}（-x）dx+{∫}_{0}^{1}（{x}^{2}+{3}^{x}）dx$=（-$\frac{1}{2}{x}^{2}$）|${\;}_{-1}^{0}$+（$\frac{1}{3}{x}^{3}+\frac{{3}^{x}}{ln3}$）|${\;}_{0}^{1}$=$\frac{1}{2}$+$\frac{1}{3}$+$\frac{3}{ln3}$-$\frac{1}{ln3}$=$\frac{5}{6}+\frac{2}{ln3}$；
故答案為：$\frac{5}{6}+\frac{2}{ln3}$．

點(diǎn)評(píng) 本題考查了定積分的運(yùn)算法則的運(yùn)用；關(guān)鍵是根據(jù)已知分段函數(shù)將定積分寫(xiě)成兩個(gè)定積分的和的形式．