Đáp án:
$\begin{array}{l}
\int\limits_1^3 {f\left( x \right)dx} = a;\int\limits_1^3 {g\left( x \right)dx} = b\\
Do:\left\{ \begin{array}{l}
\int\limits_1^3 {\left[ {f\left( x \right) + 3g\left( x \right)} \right]dx = 10} \\
\int\limits_1^3 {\left[ {2f\left( x \right) - g\left( x \right)} \right]dx} = 6
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
\int\limits_1^3 {f\left( x \right)dx + 3\int\limits_1^3 {g\left( x \right)dx} = 10} \\
2\int\limits_1^3 {f\left( x \right)dx} - \int\limits_1^3 {g\left( x \right)dx} = 6
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
a + 3b = 10\\
2a - b = 6
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
a = 4\\
b = 2
\end{array} \right.\\
\int\limits_1^3 {\left[ {f\left( x \right) + g\left( x \right)} \right]dx} \\
= \int\limits_1^3 {f\left( x \right)dx} + \int\limits_1^3 {g\left( x \right)dx} \\
= a + b\\
= 6\\
12)\\
Đặt:\int {f\left( x \right)dx} = F\left( x \right)\\
\int\limits_0^{10} {f\left( x \right)dx} = F\left( {10} \right) - F\left( 0 \right) = 10\\
\int\limits_2^5 {f\left( x \right)dx} = F\left( 5 \right) - F\left( 2 \right) = 8\\
\Rightarrow \int\limits_0^2 {f\left( x \right)dx} + \int\limits_5^{10} {f\left( x \right)dx} \\
= F\left( 2 \right) - F\left( 0 \right) + F\left( {10} \right) - F\left( 5 \right)\\
= \left( {F\left( {10} \right) - F\left( 0 \right)} \right) - \left( {F\left( 5 \right) - F\left( 2 \right)} \right)\\
= 10 - 8\\
= 2
\end{array}$
