Question

18．設(shè)f（x）=$\left\{\begin{array}{l}{{x}^{2}（x≤0）}\\{cosx-1（x＞0）}\end{array}\right.$，試求${∫}_{-1}^{\frac{π}{2}}$f（x）dx．

Answer 1

分析 先根據(jù)分段函數(shù)得到${∫}_{-1}^{\frac{π}{2}}$f（x）dx=${∫}_{0}^{\frac{π}{2}}$（cosx-1）dx+${∫}_{-1}^{0}$x²dx，再根據(jù)定積分的計(jì)算法則計(jì)算即可．

解答 解：設(shè)f（x）=$\left\{\begin{array}{l}{{x}^{2}（x≤0）}\\{cosx-1（x＞0）}\end{array}\right.$，
${∫}_{-1}^{\frac{π}{2}}$f（x）dx=${∫}_{0}^{\frac{π}{2}}$（cosx-1）dx+${∫}_{-1}^{0}$x²dx=（sinx-x）|${\;}_{0}^{\frac{π}{2}}$+$\frac{1}{3}{x}^{3}$|${\;}_{-1}^{0}$=1-$\frac{π}{2}$+$\frac{1}{3}$=$\frac{4}{3}$+$\frac{π}{2}$．

點(diǎn)評 本題考查了分段函數(shù)和定積分的計(jì)算，屬于基礎(chǔ)題．