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Câu hỏi: Cho hai tích phân $\int\limits_{-2}^{5}{f\left( x \right)\text{d}}x=8$ và $\int\limits_{5}^{-2}{g\left( x \right)\text{d}}x=3$. Tính $I=\int\limits_{-2}^{5}{\left[ f\left( x \right)-4g\left( x \right)-1 \right]\text{d}}x$
A. $13$.
B. $27$.
C. $-11$.
D. $19$.
$=\int\limits_{-2}^{5}{f\left( x \right)\text{d}x}+4\int\limits_{5}^{-2}{g\left( x \right)}\text{d}x-\int\limits_{-2}^{5}{\text{d}x}$ $=8+4.3-x\left| \begin{aligned}
& 5 \\
& -2 \\
\end{aligned} \right. $ $ =8+4.3-7 $ $ =13$.
A. $13$.
B. $27$.
C. $-11$.
D. $19$.
$I=\int\limits_{-2}^{5}{\left[ f\left( x \right)-4g\left( x \right)-1 \right]\text{d}}x$ $=\int\limits_{-2}^{5}{f\left( x \right)\text{d}x}-\int\limits_{-2}^{5}{4g\left( x \right)}\text{d}x-\int\limits_{-2}^{5}{\text{d}x}$ $=\int\limits_{-2}^{5}{f\left( x \right)\text{d}x}-4\int\limits_{-2}^{5}{g\left( x \right)}\text{d}x-\int\limits_{-2}^{5}{\text{d}x}$ $=\int\limits_{-2}^{5}{f\left( x \right)\text{d}x}+4\int\limits_{5}^{-2}{g\left( x \right)}\text{d}x-\int\limits_{-2}^{5}{\text{d}x}$ $=8+4.3-x\left| \begin{aligned}
& 5 \\
& -2 \\
\end{aligned} \right. $ $ =8+4.3-7 $ $ =13$.
Đáp án A.