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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. $I=-11$.
B. $I=13$.
C. $I=27$.
D. $I=3$.
Ta có:
$I=\int\limits_{-2}^{5}{\left[ f\left( x \right)-4g\left( x \right)-1 \right]\text{d}x}$ $=\int\limits_{-2}^{5}{f\left( x \right)\text{d}x}+4\int\limits_{5}^{-2}{g\left( x \right)\text{d}x}-\left. x \right|_{-2}^{5}$ $=8+4.3-\left( 5+2 \right)=13$ .
A. $I=-11$.
B. $I=13$.
C. $I=27$.
D. $I=3$.
Ta có:
$I=\int\limits_{-2}^{5}{\left[ f\left( x \right)-4g\left( x \right)-1 \right]\text{d}x}$ $=\int\limits_{-2}^{5}{f\left( x \right)\text{d}x}+4\int\limits_{5}^{-2}{g\left( x \right)\text{d}x}-\left. x \right|_{-2}^{5}$ $=8+4.3-\left( 5+2 \right)=13$ .
Đáp án B.