Cho hàm số \(y=f\left( x \right)=1, y=g\left( x \right)=\left| x \right|\). Giá trị \(I=\int\limits_{-1}^{2}{\min \left\{ f\left( x \right);g\left( x \right) \right\}}\text{d}x\)

* Hướng dẫn giải

Xét bất phương trình \( \left| x \right|>1\Leftrightarrow \left[ \begin{align} & x>1 \\ & x<-1 \\ \end{align} \right.\)

Vậy \(\min \left\{ 1;\left| x \right| \right\}=1\) khi 1<x hoặc x<-1

\(\min \left\{ 1;\left| x \right| \right\}=\left| x \right|\) khi -1<x<1

Xét \(I=\int\limits_{-1}^{2}{\min \left\{ f\left( x \right);g\left( x \right) \right\}}\text{d}x=\int\limits_{-1}^{2}{\min \left\{ 1;\left| x \right| \right\}}\text{d}x=\int\limits_{-1}^{1}{\min \left\{ 1;\left| x \right| \right\}}\text{d}x+\int\limits_{1}^{2}{\min \left\{ 1;\left| x \right| \right\}}\text{d}x\)

\(I=\int\limits_{-1}^{1}{\left| x \right|}\text{d}x+\int\limits_{1}^{2}{\text{d}x}=\int\limits_{-1}^{0}{-x}\text{d}x+\int\limits_{0}^{1}{x}\text{d}x+\int\limits_{1}^{2}{\text{d}x}=\left. \frac{-{{x}^{2}}}{2} \right|_{-1}^{0}+\left. \frac{{{x}^{2}}}{2} \right|_{0}^{1}+\left. x \right|_{1}^{2}=2.\)