Question

13．已知f（x）=$\left\{\begin{array}{l}{x+1，x≤1}\\{\frac{1}{2}{x}^{2}，x＞1}\end{array}\right.$，求${∫}_{0}^{2}$f（x）dx．

Answer 1

分析 根据定积分的计算法则计算即可．

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+${∫}_{1}^{2}$（$\frac{1}{2}$x²）dx=（$\frac{1}{2}$x²+x）|${\;}_{0}^{1}$+$\frac{1}{6}$x³|${\;}_{1}^{2}$=$\frac{1}{2}$+1+$\frac{1}{6}$×（8-1）=$\frac{8}{3}$

点评 本题考查了定积分的计算，关键是求出原函数，属于基础题．