B. Biết \(\int\limits_1^2 {f\left( x \right){\mkern 1mu} } {\rm{d}}x =  - 1\) và \(\int\limits_1^2 {g\left( x \right){\mkern 1mu} } {\rm{d}}x = 3\), khi đó \(\int\limits_2^1 {\left[ {f\left( x \right) - g\left( x \right)} \right]{\mkern 1mu} } {\rm{d}}x = 5\)   

Biết \(\int\limits_1^2 {f\left( x \right){\mkern 1mu} } {\rm{d}}x =  - 1\) và \(\int\limits_1^2 {g\left( x \right){\mkern 1mu} } {\rm{d}}x = 3\), khi đó \(\int\limits_2^1 {\left[ {f\left( x \right) - g\left( x \right)} \right]{\mkern 1mu} } {\rm{d}}x = 5\)         

           Ta có: \(\int\limits_1^2 {f\left( x \right){\mkern 1mu} } {\rm{d}}x =  - 1 \Leftrightarrow \int\limits_2^1 {f\left( x \right){\mkern 1mu} } {\rm{d}}x = 1;\)     \(\int\limits_1^2 {g\left( x \right){\mkern 1mu} } {\rm{d}}x = 3 \Leftrightarrow \int\limits_2^1 {g\left( x \right){\mkern 1mu} } {\rm{d}}x =  - 3\)

           \(\int\limits_2^1 {\left[ {f\left( x \right) - g\left( x \right)} \right]\,} {\rm{d}}x = \int\limits_2^1 {f\left( x \right)\,} {\rm{d}}x - \int\limits_2^1 {g\left( x \right)\,} {\rm{d}}x = 1 - \left( { - 3} \right) = 4\).