Question

1．$f（x）=\left\{\begin{array}{l}x+1，x≤1\\ \frac{1}{2}{x^2}，x＞1\end{array}\right.$，求$\int_{\;0}^{\;2}{f（x）dx}$=$\frac{8}{3}$．

Answer 1

分析 由$\int_{\;0}^{\;2}{f（x）dx}$=${∫}_{0}^{1}$（x+1）dx+${∫}_{1}^{2}$$\frac{1}{2}{x}^{2}$dx，根據(jù)定積分的計(jì)算法則計(jì)算即可．

解答 解：$f（x）=\left\{\begin{array}{l}x+1，x≤1\\ \frac{1}{2}{x^2}，x＞1\end{array}\right.$，
則$\int_{\;0}^{\;2}{f（x）dx}$=${∫}_{0}^{1}$（x+1）dx+${∫}_{1}^{2}$$\frac{1}{2}{x}^{2}$dx=（$\frac{1}{2}$x²+x）|${\;}_{0}^{1}$+$\frac{1}{6}{x}^{3}$|${\;}_{1}^{2}$=$\frac{1}{2}$+1+$\frac{1}{6}$（2³-1）=$\frac{8}{3}$，
故答案為：$\frac{8}{3}$．

點(diǎn)評(píng) 本題考查了定積分的計(jì)算，關(guān)鍵是轉(zhuǎn)化，屬于基礎(chǔ)題．