Question

8．求定积分：∫${\;}_{0}^{2}$f（x）dx，其中f（x）=$\left\{\begin{array}{l}{x+1，x≤1}\\{\frac{1}{2}{x}^{2}，x＞1}\end{array}\right.$．

Answer 1

分析 直接利用定积分的求解方法求解即可．

解答 解：f（x）=$\left\{\begin{array}{l}{x+1，x≤1}\\{\frac{1}{2}{x}^{2}，x＞1}\end{array}\right.$．
∫${\;}_{0}^{2}$f（x）dx=${∫}_{0}^{1}（x+1）dx+{∫}_{1}^{2}（\frac{1}{2}{x}^{2}）dx$=（$\frac{1}{2}{x}^{2}$+x）${|}_{0}^{1}$+$\frac{1}{6}{x}^{3}{|}_{1}^{2}$
=$\frac{3}{2}$+$\frac{8}{6}$$-\frac{1}{6}$
=$\frac{8}{3}$．

点评 本题考查定积分的求法，考查计算能力．