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Câu hỏi: Cho $\int_{-1}^{2}{f\left( x \right)dx}=2$ và $\int_{-1}^{2}{g\left( x \right)dx}=-1$. Tính $I=\int_{-1}^{2}{\left[ x+2f\left( x \right)+3g\left( x \right) \right]dx}$
A. $I=\dfrac{11}{2}$
B. $I=\dfrac{7}{2}$
C. $I=\dfrac{17}{2}$
D. $I=\dfrac{5}{2}$
A. $I=\dfrac{11}{2}$
B. $I=\dfrac{7}{2}$
C. $I=\dfrac{17}{2}$
D. $I=\dfrac{5}{2}$
$I=\int\limits_{-1}^{2}{\left[ x+2f\left( x \right)+3g\left( x \right) \right]dx}=\int\limits_{-1}^{2}{xdx}+2\int\limits_{-1}^{2}{f\left( x \right)dx}+3\int\limits_{-1}^{2}{g\left( x \right)dx}$
$=\dfrac{{{x}^{2}}}{2}\left| \begin{aligned}
& ^{2} \\
& _{-1} \\
\end{aligned} \right.+2.2+3.\left( -1 \right)=\left( 2-\dfrac{1}{2} \right)+1=\dfrac{5}{2}$
$=\dfrac{{{x}^{2}}}{2}\left| \begin{aligned}
& ^{2} \\
& _{-1} \\
\end{aligned} \right.+2.2+3.\left( -1 \right)=\left( 2-\dfrac{1}{2} \right)+1=\dfrac{5}{2}$
Đáp án D.