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Câu hỏi: Cho $\int\limits_{-2}^{5}{f\left( x \right)dx}=8$ và $\int\limits_{-2}^{5}{g\left( x \right)dx}=-3.$ Tính $I=\int\limits_{-2}^{5}{\left[ f\left( x \right)-4g\left( x \right)-1 \right]dx}.$
A. $I=3$
B. $I=13$
C. $I=-11$
D. $I=27$
A. $I=3$
B. $I=13$
C. $I=-11$
D. $I=27$
Phương pháp:
Sử dụng tính chất tích phân: $\int\limits_{a}^{b}{\left[ f\left( x \right)+g\left( x \right) \right]dx}=\int\limits_{a}^{b}{f\left( x \right)dx}+\int\limits_{b}^{b}{g\left( x \right)dx},\int\limits_{a}^{b}{kf\left( x \right)dx}=k\int\limits_{a}^{b}{f\left( x \right)dx}\left( k\ne 0 \right).$
Cách giải:
Ta có:
$I=\int\limits_{-2}^{5}{\left[ f\left( x \right)-4g\left( x \right)-1 \right]dx}$
$=\int\limits_{-2}^{5}{f\left( x \right)dx}-4\int\limits_{-2}^{5}{g\left( x \right)dx}-\int\limits_{-2}^{5}{dx}$
$=8-4.\left( -3 \right)-x\left| \begin{aligned}
& 5 \\
& -2 \\
\end{aligned} \right.$
$=20-\left( 5-\left( -2 \right) \right)=13.$
Sử dụng tính chất tích phân: $\int\limits_{a}^{b}{\left[ f\left( x \right)+g\left( x \right) \right]dx}=\int\limits_{a}^{b}{f\left( x \right)dx}+\int\limits_{b}^{b}{g\left( x \right)dx},\int\limits_{a}^{b}{kf\left( x \right)dx}=k\int\limits_{a}^{b}{f\left( x \right)dx}\left( k\ne 0 \right).$
Cách giải:
Ta có:
$I=\int\limits_{-2}^{5}{\left[ f\left( x \right)-4g\left( x \right)-1 \right]dx}$
$=\int\limits_{-2}^{5}{f\left( x \right)dx}-4\int\limits_{-2}^{5}{g\left( x \right)dx}-\int\limits_{-2}^{5}{dx}$
$=8-4.\left( -3 \right)-x\left| \begin{aligned}
& 5 \\
& -2 \\
\end{aligned} \right.$
$=20-\left( 5-\left( -2 \right) \right)=13.$
Đáp án B.