Ta có
\(\int\limits_0^2 {f\left( x \right)dx} = \int\limits_0^2 {2xdx} = 2\int\limits_0^2 {xdx} = 2\left. {\left( {\frac{{{x^2}}}{2}} \right)} \right|_0^2 = 2\left( {\frac{{{2^2}}}{2} - \frac{{{0^2}}}{2}} \right) = 4\)
\(\int\limits_0^1 {f\left( x \right)dx} + \int\limits_1^2 {f\left( x \right)dx} = \int\limits_0^1 {2xdx} + \int\limits_1^2 {2xdx} = 2\left( {\int\limits_0^1 {xdx} + \int\limits_1^2 {xdx} } \right) = 2\left[ {\left. {\left( {\frac{{{x^2}}}{2}} \right)} \right|_0^1 + \left. {\left( {\frac{{{x^2}}}{2}} \right)} \right|_1^2} \right]\)\( = 2\left[ {\left( {\frac{{{1^2}}}{2} - \frac{{{0^2}}}{2}} \right) + \left( {\frac{{{2^2}}}{2} - \frac{{{1^2}}}{2}} \right)} \right] = 4\)
Vậy \(\int\limits_0^2 {f\left( x \right)dx} = \int\limits_0^1 {f\left( x \right)dx} + \int\limits_1^2 {f\left( x \right)dx} \)