Chọn B

\(\int\limits_0^2 {\left[ {\frac{1}{2}f\left( x \right) + 2} \right]} {\rm{d}}x = \frac{1}{2}\int\limits_0^2 {f\left( x \right)} {\rm{d}}x + \int\limits_0^2 2 {\rm{d}}x = 2 + 4 = 6\).

Chọn A

\(\int\limits_5^{ - 1} {f\left( x \right){\rm{d}}x} = - \int\limits_{ - 1}^5 {f\left( x \right){\rm{d}}x} = 3\).

Chọn A

Ta có \(\int\limits_0^3 {\left[ {\frac{1}{3}f\left( x \right) + 2} \right]{\rm{d}}x} = \frac{1}{3}\int\limits_0^3 {f\left( x \right)dx + \int\limits_0^3 {2dx = \frac{1}{3} \cdot 6 + 6 = 8} } \).

Chọn B

Ta có \(\int\limits_{ - 1}^5 {f\left( x \right){\rm{d}}x = } \int\limits_{ - 1}^2 {f\left( x \right){\rm{d}}x + } \int\limits_2^5 {f\left( x \right){\rm{d}}x = 2 - 5 = - 3} \).

Chọn B

Ta có: \(\int\limits_{ - 1}^5 {f\left( x \right){\rm{d}}x} = \int\limits_{ - 1}^2 {f\left( x \right){\rm{d}}x} + \int\limits_2^5 {f\left( x \right){\rm{d}}x} = 2 + \left( { - 5} \right) = - 3\).

Chọn D

Ta có \(\int\limits_0^3 {\left[ {\frac{1}{3}f\left( x \right) + 2} \right]{\rm{d}}x} = \frac{1}{3}\int\limits_0^3 {f\left( x \right){\rm{d}}x} + \int\limits_0^3 {{\rm{2d}}x} = 2 + 6 = 8\).

Chọn B

Ta có \(\int\limits_0^2 {\left[ {4x - f\left( x \right)} \right]{\rm{d}}x} \)\( = \)\(\int\limits_0^2 {4x{\rm{d}}x} - \int\limits_0^2 {f\left( x \right){\rm{d}}x = } \left. {2{x^2}} \right|_0^2 - 3 = 8 - 3 = 5\)

 Chọn C

Ta có \(\int\limits_0^2 {\left[ {2x - f\left( x \right)} \right]{\rm{d}}x} = \int\limits_0^2 {2x{\rm{d}}x} - \int\limits_0^2 {f\left( x \right){\rm{d}}x} \)\( = \left. {{x^2}} \right|_0^2 - 3 = 4 - 3 = 1\).

Chọn B

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

 Chọn D

Ta có: \(\int\limits_0^1 {2f\left( x \right){\rm{d}}x} = 2\int\limits_0^1 {f\left( x \right){\rm{d}}x} = 2.4 = 8\).

Chọn C

Ta có: \(\int\limits_1^3 {2f\left( x \right){\rm{d}}x} = 2\int\limits_1^3 {f\left( x \right){\rm{d}}x} = 2.3 = 6\).

Chọn A

Ta có: \(\int\limits_1^2 {\left[ {2 + f\left( x \right)} \right]{\rm{d}}x} = \left( {2x + {x^2}} \right)\left| \begin{array}{l}2\\1\end{array} \right. = 8 - 3 = 5\)

Chọn D

Ta có ${\int }_1^{5}3f\left(x\right)dx=3{\int }_1^{5}f\left(x\right)dx=3.4=12$

Chọn C

Ta có  ${\int }_1^2(2+f(x\left)\right)dx={\int }_1^{2}f\left(x\right)dx=2x\left\{\begin{array}{l}2\\ 1\end{array}+F\left(x\right)\left\{\begin{array}{l}2\\ 1\end{array}=2x\left\{\begin{array}{l}2\\ 1\end{array}+x^3\left\{\begin{array}{l}2\\ 1\end{array}=9\right.\right.\right.\right.$

Chọn B

Ta có : \(\int\limits_1^2 {3f\left( x \right)} dx = 3\int\limits_1^2 {f\left( x \right)} dx\)\( = 3.2 = 6\).

Chọn D

Ta có \(\int\limits_1^3 {\left[ {1 + f(x)} \right]} {\rm{d}}x = \left[ {x + F(x)} \right]\left| {\mathop {}\limits_1^3 } \right. = \left[ {x + {x^3})} \right]\left| {\mathop {}\limits_1^3 } \right. = 30 - 2 = 28\).

Chọn C

Ta có : \(\int\limits_2^3 {2f\left( x \right){\rm{d}}x} = 2\int\limits_2^3 {f\left( x \right){\rm{d}}x} = 12.\).

Chọn A

Ta có \(\int\limits_1^3 {\left[ {1 + f(x)} \right]dx} = \left. {\left( {x + F\left( x \right)} \right)} \right|_1^3 = \left. {\left( {x + {x^2}} \right)} \right|_1^3 = 12 - 2 = 10.\)

Chọn B

Ta có \[\int\limits_2^3 {\left[ {f\left( x \right) - g\left( x \right)} \right]dx} = \int\limits_2^3 {f\left( x \right)dx} - \int\limits_2^3 {g\left( x \right)dx} = 4 - 1 = 3\]

Chọn A

Ta có

\[\int\limits_0^1 {\left[ {f\left( x \right) + 2{\rm{x}}} \right]} d{\rm{x = 2}} \Leftrightarrow \int\limits_0^1 {f\left( x \right)} d{\rm{x + }}\int\limits_0^1 {2{\rm{x}}} d{\rm{x = 2}}\]\[\, \Leftrightarrow \int\limits_0^1 {f\left( x \right)} d{\rm{x}} = 2 - \left. {{x^2}} \right|_0^1\]\[\, \Leftrightarrow \int\limits_0^1 {f\left( x \right)} d{\rm{x}} = 2 - 1\] \[\, \Leftrightarrow \int\limits_0^1 {f\left( x \right)} d{\rm{x}} = 1\]

Chọn A

Ta có: \(\int\limits_2^3 {\left[ {f\left( x \right) + g\left( x \right)} \right]} dx = \int\limits_2^3 {f\left( x \right)dx} + \int\limits_2^3 {g\left( x \right)dx} = 4\).

Chọn D

Ta có \(\int\limits_0^1 {\left[ {f\left( x \right) + 2x} \right]} dx = 3 \Leftrightarrow \int\limits_0^1 {f\left( x \right)} dx + 2\int\limits_0^1 x dx = 3 \Leftrightarrow \int\limits_0^1 {f\left( x \right)} dx + 2.\frac{{{x^2}}}{2}\left| {\begin{array}{*{20}{c}}1\\0\end{array}} \right. = 3\).

Suy ra \(\int\limits_0^1 {f\left( x \right){\rm{d}}x} = 3 - {x^2}\left| {\begin{array}{*{20}{c}}1\\0\end{array}} \right. = 3 - \left( {1 - 0} \right) = 2\).

Chọn B

 Ta có: \(\int\limits_1^2 {\left[ {f\left( x \right) - g\left( x \right)} \right]} \,{\rm{d}}x = \int\limits_1^2 {f\left( x \right)} \,{\rm{d}}x - \int\limits_1^2 {g\left( x \right){\rm{d}}x} = 3 - 2 = 1\).