Đáp án:
$\begin{array}{l}
13)I = \int\limits_2^6 {\left[ {3f\left( x \right) - 4x} \right]dx} \\
= 3\int\limits_2^6 {f\left( x \right)dx} - 4\int\limits_2^6 x dx\\
= 3.\left( {\int\limits_2^{ - 1} {f\left( x \right)dx} + \int\limits_{ - 1}^6 {f\left( x \right)dx} } \right) - 4\left( {\frac{{{x^2}}}{2}} \right)_2^6\\
= 3.\left( {4 + 5} \right) - 4.\left( {\frac{{{6^2}}}{2} - \frac{{{2^2}}}{2}} \right)\\
= 3.9 - 4.34\\
= 27 - 136\\
= - 109\\
14)\int\limits_2^3 {f\left( x \right)dx} \\
= \int\limits_2^0 {f\left( x \right)dx} + \int\limits_0^3 {f\left( x \right)dx} \\
= 5 + 3 = 8\\
15)\int\limits_1^3 {\left[ {x - 2f\left( x \right)} \right]dx} \\
= \int\limits_1^3 {xdx} - 2\int\limits_1^3 {f\left( x \right)dx} \\
= \left( {\frac{{{x^2}}}{2}} \right)_1^3 + 2.\int\limits_3^1 {f\left( x \right)dx} \\
= \left( {\frac{{{3^2}}}{2} - \frac{{{1^2}}}{2}} \right) + 2.\left( { - 4} \right)\\
= 4 - 8\\
= - 4\\
16)\int\limits_{ - 2}^2 {\left[ {f\left( x \right) - 2x + 1} \right]dx} \\
= \int\limits_{ - 2}^2 {f\left( x \right)} dx + \int\limits_{ - 2}^2 {\left( { - 2x + 1} \right)dx} \\
= 0 + \left( { - {x^2} + x} \right)_{ - 2}^2\\
= \left( { - {2^2} + 2 + {{\left( { - 2} \right)}^2} + 2} \right)\\
= 4
\end{array}$
