Cho \(\int\limits_{ - 3}^0 {f\left( x \right){\rm{d}}x} = - 4\) và \(\int\limits_{ - 3}^0 {g\left( x \right){\rm{d}}x} = - 3\).
Ta có: \(\int\limits_{ - 3}^0 {\left[ {f\left( x \right) + g\left( x \right)} \right]{\rm{d}}x} = \int\limits_{ - 3}^0 {f\left( x \right){\rm{d}}x} + \int\limits_{ - 3}^0 {g\left( x \right){\rm{d}}x} = \left( { - 4} \right) + \left( { - 3} \right) = - 7\).
\[\int\limits_{ - 3}^0 {\left[ {f\left( x \right) - g\left( x \right)} \right]{\rm{d}}x} \]\( = \int\limits_{ - 3}^0 {f\left( x \right){\rm{d}}x} - \int\limits_{ - 3}^0 {g\left( x \right){\rm{d}}x} = \left( { - 4} \right) - \left( { - 3} \right) = - 1\).
\(\int\limits_{ - 3}^0 {\left[ { - 3f\left( x \right)} \right]{\rm{d}}x} = \left( { - 3} \right) \cdot \int\limits_{ - 3}^0 {f\left( x \right){\rm{d}}x} = \left( { - 3} \right) \cdot \left( { - 4} \right) = 12\).
\[\int\limits_{ - 3}^0 {\left[ {2f\left( x \right) + 3g\left( x \right)} \right]{\rm{d}}x} = 2\int\limits_{ - 3}^0 {f\left( x \right){\rm{d}}x} + 3\int\limits_{ - 3}^0 {g\left( x \right){\rm{d}}x} \]\( = 2 \cdot \left( { - 4} \right) + 3 \cdot \left( { - 3} \right) = - 17\).
Đáp án: a) Đúng, b) Sai, c) Đúng, d) Sai.