Google
Câu hỏi: Cho ${\int\limits_{-1}^{2}{f\left( x \right)\text{d}x}=3, \int\limits_{-1}^{2}{g\left( x \right)\text{d}x}=-1}$. Khi đó ${I=\int\limits_{-1}^{2}{\left[ x+2f\left( x \right)-3g\left( x \right) \right]\text{d}x}}$ bằng
A. $10$.
B. $\dfrac{21}{2}$.
C. $\dfrac{19}{2}$.
D. $\dfrac{17}{2}$.
A. $10$.
B. $\dfrac{21}{2}$.
C. $\dfrac{19}{2}$.
D. $\dfrac{17}{2}$.
Ta có:
$I=\int\limits_{-1}^{2}{\left[ x+2f\left( x \right)-3g\left( x \right) \right]\text{d}x}=\int\limits_{-1}^{2}{x\text{d}x}+2\int\limits_{-1}^{2}{f\left( x \right)\text{d}x}-3\int\limits_{-1}^{2}{g\left( x \right)\text{d}x}=\dfrac{3}{2}+2.3-3.\left( -1 \right)=\dfrac{21}{2}$.
$I=\int\limits_{-1}^{2}{\left[ x+2f\left( x \right)-3g\left( x \right) \right]\text{d}x}=\int\limits_{-1}^{2}{x\text{d}x}+2\int\limits_{-1}^{2}{f\left( x \right)\text{d}x}-3\int\limits_{-1}^{2}{g\left( x \right)\text{d}x}=\dfrac{3}{2}+2.3-3.\left( -1 \right)=\dfrac{21}{2}$.
Đáp án B.