Question

11．已知函數(shù)f（x）=$\left\{\begin{array}{l}{x+1，（-1≤x≤0）}\\{cosx，（0＜x≤\frac{π}{2}）}\end{array}\right.$，則${∫}_{-1}^{\frac{π}{2}}$f（x）dx=（　�。�

• A． $\frac{1}{2}$
• B． 1
• C． 2
• D． $\frac{3}{2}$

Answer 1

分析 由題意得到${∫}_{-1}^{\frac{π}{2}}$f（x）dx=${∫}_{-1}^{0}$（x+1）dx+${∫}_{0}^{\frac{π}{2}}$cosxdx，解得即可．

解答 解：f（x）=$\left\{\begin{array}{l}{x+1，（-1≤x≤0）}\\{cosx，（0＜x≤\frac{π}{2}）}\end{array}\right.$，
則${∫}_{-1}^{\frac{π}{2}}$f（x）dx=${∫}_{-1}^{0}$（x+1）dx+${∫}_{0}^{\frac{π}{2}}$cosxdx=（$\frac{1}{2}$x²+x）|${\;}_{-1}^{0}$+sinx|${\;}_{0}^{\frac{π}{2}}$=（$\frac{1}{2}$-1）+sin$\frac{π}{2}$-sin0=-$\frac{1}{2}$+1=$\frac{1}{2}$，
故選：A．

點(diǎn)評(píng) 本題考查了定積分的計(jì)算，關(guān)鍵是求出原函數(shù)，屬于基礎(chǔ)題．