分段積分試題解析及答案

一、單項選擇題（每題2分，共10題）1.函數(shù)\(f(x)=\begin{cases}x,&0\leqx\lt1\\2,&1\leqx\leq2\end{cases}\)在區(qū)間\([0,2]\)上的定積分是（）A.\(\frac{3}{2}\)B.\(\frac{5}{2}\)C.\(3\)D.\(4\)2.已知\(f(x)=\begin{cases}e^x,&x\lt0\\x+1,&x\geq0\end{cases}\)，則\(\int_{-1}^1{f(x)dx}\)的值為（）A.\(e+\frac{1}{2}\)B.\(e-\frac{1}{2}\)C.\(\frac{1}{e}+\frac{1}{2}\)D.\(\frac{1}{e}-\frac{1}{2}\)3.設\(f(x)=\begin{cases}2x,&0\leqx\leq1\\3-x,&1\ltx\leq2\end{cases}\)，則\(\int_{0}^{2}{f(x)dx}\)等于（）A.\(\frac{3}{2}\)B.\(\frac{5}{2}\)C.\(\frac{7}{2}\)D.\(\frac{9}{2}\)4.函數(shù)\(y=\begin{cases}x^2,&0\leqx\lt1\\1,&1\leqx\leq2\end{cases}\)在\([0,2]\)上的平均值是（）A.\(\frac{4}{3}\)B.\(\frac{5}{3}\)C.\(\frac{7}{3}\)D.\(\frac{8}{3}\)5.若\(f(x)=\begin{cases}1,&x\in[0,1]\\2,&x\in(1,2]\end{cases}\)，則\(\int_{0}^{2}{f(x)dx}\)的值為（）A.\(1\)B.\(2\)C.\(3\)D.\(4\)6.已知\(f(x)=\begin{cases}x^3,&x\lt0\\\sqrt{x},&x\geq0\end{cases}\)，那么\(\int_{-1}^1{f(x)dx}\)為（）A.\(\frac{1}{4}+\frac{2}{3}\)B.\(-\frac{1}{4}+\frac{2}{3}\)C.\(\frac{1}{4}-\frac{2}{3}\)D.\(-\frac{1}{4}-\frac{2}{3}\)7.函數(shù)\(f(x)=\begin{cases}2-x,&0\leqx\leq2\\0,&x\gt2\end{cases}\)在\([0,3]\)上的定積分是（）A.\(2\)B.\(3\)C.\(4\)D.\(5\)8.設\(f(x)=\begin{cases}1,&x\in[0,\pi]\\0,&x\notin[0,\pi]\end{cases}\)，則\(\int_{-\pi}^{\pi}{f(x)dx}\)等于（）A.\(0\)B.\(\pi\)C.\(2\pi\)D.\(3\pi\)9.已知\(f(x)=\begin{cases}2x+1,&0\leqx\lt1\\3,&1\leqx\leq2\end{cases}\)，則\(\int_{0}^{2}{f(x)dx}\)的值為（）A.\(4\)B.\(5\)C.\(6\)D.\(7\)10.函數(shù)\(y=\begin{cases}x+1,&0\leqx\lt1\\2,&1\leqx\leq2\end{cases}\)在\([0,2]\)上的定積分是（）A.\(\frac{5}{2}\)B.\(\frac{7}{2}\)C.\(\frac{9}{2}\)D.\(\frac{11}{2}\)答案：1.B2.C3.C4.A5.C6.B7.A8.B9.B10.B二、多項選擇題（每題2分，共10題）1.對于分段函數(shù)\(f(x)=\begin{cases}x^2,&x\in[0,1]\\2-x,&x\in(1,2]\end{cases}\)，以下說法正確的是（）A.\(f(x)\)在\([0,2]\)上連續(xù)B.\(\int_{0}^{1}{f(x)dx}=\frac{1}{3}\)C.\(\int_{1}^{2}{f(x)dx}=\frac{1}{2}\)D.\(\int_{0}^{2}{f(x)dx}=\frac{5}{6}\)2.已知分段函數(shù)\(g(x)=\begin{cases}e^x,&x\lt0\\1+x,&x\geq0\end{cases}\)，則（）A.\(g(x)\)在\(x=0\)處可導B.\(\int_{-1}^0{g(x)dx}=\frac{1}{e}-1\)C.\(\int_{0}^{1}{g(x)dx}=\frac{3}{2}\)D.\(\int_{-1}^1{g(x)dx}=\frac{1}{e}+\frac{1}{2}\)3.設分段函數(shù)\(h(x)=\begin{cases}x+1,&0\leqx\lt1\\3-x,&1\leqx\leq2\end{cases}\)，則（）A.\(h(x)\)在\(x=1\)處連續(xù)B.\(\int_{0}^{1}{h(x)dx}=\frac{3}{2}\)C.\(\int_{1}^{2}{h(x)dx}=\frac{3}{2}\)D.\(\int_{0}^{2}{h(x)dx}=3\)4.對于函數(shù)\(y=\begin{cases}2x,&0\leqx\leq1\\4-2x,&1\ltx\leq2\end{cases}\)，下列正確的是（）A.函數(shù)圖象關(guān)于\(x=1\)對稱B.\(\int_{0}^{1}{ydx}=1\)C.\(\int_{1}^{2}{ydx}=1\)D.\(\int_{0}^{2}{ydx}=2\)5.已知分段函數(shù)\(k(x)=\begin{cases}x^3,&x\in[-1,0)\\x^2,&x\in[0,1]\end{cases}\)，則（）A.\(k(x)\)在\(x=0\)處連續(xù)B.\(\int_{-1}^0{k(x)dx}=-\frac{1}{4}\)C.\(\int_{0}^{1}{k(x)dx}=\frac{1}{3}\)D.\(\int_{-1}^1{k(x)dx}=\frac{1}{12}\)6.設\(f(x)=\begin{cases}2x,&0\leqx\leqa\\2a,&a\ltx\leq2a\end{cases}\)（\(a\gt0\)），則（）A.\(\int_{0}^{a}{f(x)dx}=a^2\)B.\(\int_{a}^{2a}{f(x)dx}=2a^2\)C.\(\int_{0}^{2a}{f(x)dx}=3a^2\)D.\(f(x)\)在\(x=a\)處連續(xù)7.對于分段函數(shù)\(m(x)=\begin{cases}\sinx,&x\in[0,\pi]\\0,&x\notin[0,\pi]\end{cases}\)，有（）A.\(\int_{0}^{\pi}{m(x)dx}=2\)B.\(\int_{-\pi}^0{m(x)dx}=0\)C.\(\int_{-\pi}^{\pi}{m(x)dx}=2\)D.\(m(x)\)在\(x=\pi\)處連續(xù)8.已知\(n(x)=\begin{cases}1-x,&0\leqx\lt1\\x-1,&1\leqx\leq2\end{cases}\)，則（）A.\(n(x)\)的圖象關(guān)于直線\(x=1\)對稱B.\(\int_{0}^{1}{n(x)dx}=\frac{1}{2}\)C.\(\int_{1}^{2}{n(x)dx}=\frac{1}{2}\)D.\(\int_{0}^{2}{n(x)dx}=1\)9.設\(p(x)=\begin{cases}x^2+1,&0\leqx\lt1\\3,&1\leqx\leq2\end{cases}\)，則（）A.\(\int_{0}^{1}{p(x)dx}=\frac{4}{3}\)B.\(\int_{1}^{2}{p(x)dx}=3\)C.\(\int_{0}^{2}{p(x)dx}=\frac{13}{3}\)D.\(p(x)\)在\(x=1\)處不連續(xù)10.對于分段函數(shù)\(q(x)=\begin{cases}2^x,&x\in[0,1]\\1,&x\in(1,2]\end{cases}\)，以下正確的是（）A.\(\int_{0}^{1}{q(x)dx}=\frac{1}{\ln2}\)B.\(\int_{1}^{2}{q(x)dx}=1\)C.\(\int_{0}^{2}{q(x)dx}=\frac{1}{\ln2}+1\)D.\(q(x)\)在\(x=1\)處連續(xù)答案：1.BCD2.BCD3.ACD4.ABCD5.ABCD6.ABC7.ABC8.ABCD9.ACD10.BC三、判斷題（每題2分，共10題）1.分段函數(shù)\(f(x)=\begin{cases}x+1,&x\lt0\\x-1,&x\geq0\end{cases}\)在\(x=0\)處連續(xù)。（）2.若\(g(x)=\begin{cases}x^2,&0\leqx\leq1\\2-x,&1\ltx\leq2\end{cases}\)，則\(\int_{0}^{2}{g(x)dx}=\frac{5}{6}\)。（）3.分段函數(shù)\(h(x)=\begin{cases}e^x,&x\lt0\\1+x,&x\geq0\end{cases}\)在\(x=0\)處可導。（）4.對于\(y=\begin{cases}2x,&0\leqx\leq1\\4-2x,&1\ltx\leq2\end{cases}\)，\(\int_{0}^{2}{ydx}=2\)。（）5.設\(k(x)=\begin{cases}x^3,&x\in[-1,0)\\x^2,&x\in[0,1]\end{cases}\)，\(k(x)\)在\(x=0\)處連續(xù)。（）6.已知\(f(x)=\begin{cases}2x,&0\leqx\leqa\\2a,&a\ltx\leq2a\end{cases}\)（\(a\gt0\)），\(\int_{0}^{2a}{f(x)dx}=3a^2\)。（）7.分段函數(shù)\(m(x)=\begin{cases}\sinx,&x\in[0,\pi]\\0,&x\notin[0,\pi]\end{cases}\)在\(x=\pi\)處連續(xù)。（）8.對于\(n(x)=\begin{cases}1-x,&0\leqx\lt1\\x-1,&1\leqx\leq2\end{cases}\)，\(\int_{0}^{2}{n(x)dx}=1\)。（）9.設\(p(x)=\begin{cases}x^2+1,&0\leqx\lt1\\3,&1\leqx\leq2\end{cases}\)，\(p(x)\)在\(x=1\)處不連續(xù)。（）10.已知\(q(x)=\begin{cases}2^x,&x\in[0,1]\\1,&x\in(1,2]\end{cases}\)，\(\int_{0}^{2}{q(x)dx}=\frac{1}{\ln2}+1\)。（）答案：1.×2.√3.×4.√5.√6.√7.×8.√9.√10.√四、簡答題（每題5分，共4題）1.求分段函數(shù)\(f(x)=\begin{cases}x,&0\leqx\lt1\\2,&1\leqx\leq2\end{cases}\)在區(qū)間\([0,2]\)上的定積分。答案：\(\int_{0}^{2}{f(x)dx}=\int_{0}^{1}{xdx}+\int_{1}^{2}{2dx}\)。\(\int_{0}^{1}{xdx}=\frac{1}{2}x^2\big|_0^1=\frac{1}{2}\)，\(\int_{1}^{2}{2dx}=2x\big|_1^2=2\)，所以定積分值為\(\frac{1}{2}+2=\frac{5}{2}\)。2.已知\(g(x)=\begin{cases}e^x,&x\lt0\\1+x,&