Cho hàm số f( x ),g( x ) liên tục trên R. Các mệnh đề sau đây đúng hay sai?

A. Biết \[\int\limits_1^2 {f\left( x \right)dx}  = 2\]. Giá trị của \[\int\limits_2^1 {3f\left( x \right)dx}  =  - 6\].

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

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

C-Đúng

Nếu \(\int\limits_1^2 {f\left( x \right){\rm{d}}x}  =  - 2\) và \(\int\limits_2^3 {f\left( x \right){\rm{d}}x}  = 1\) thì \(\int\limits_1^3 {f\left( x \right){\rm{d}}x}  =  - 1\).

           Ta có \(\int\limits_1^3 {f\left( x \right){\rm{d}}x}  = \int\limits_1^2 {f\left( x \right){\rm{d}}x}  + \int\limits_2^3 {f\left( x \right){\rm{d}}x}  =  - 2 + 1 =  - 1\).

D-Đúng 

Nếu \(\int\limits_0^2 {\left( {f\left( x \right) + 3{x^2}} \right){\rm{d}}x}  = 10\) thì \(\int\limits_0^2 {f\left( x \right){\rm{d}}x}  = 2\).

Ta có:

\(\,\,\,\,\,\,\int\limits_0^2 {\left( {f\left( x \right) + 3{x^2}} \right){\rm{d}}x}  = 10\) \( \Leftrightarrow \int\limits_0^2 {f\left( x \right)} {\rm{d}}x + \int\limits_0^2 {3{x^2}} {\rm{d}}x = 10\) \( \Leftrightarrow \int\limits_0^2 {f\left( x \right)} {\rm{d}}x = 10 - \int\limits_0^2 {3{x^2}} {\rm{d}}x\)

\( \Leftrightarrow \int\limits_0^2 {f\left( x \right)} {\rm{d}}x = 10 - {x^3}\left| \begin{array}{l}2\\0\end{array} \right.\,\) \( \Leftrightarrow \int\limits_0^2 {f\left( x \right)} {\rm{d}}x = 10 - 8 = 2\).