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Câu hỏi: Cho $f\left( x \right),g\left( x \right)$ là hai hàm số liên tục trên $\left[ 0;2 \right]$ thỏa mãn điều kiện $\int\limits_{0}^{2}{\left[ f\left( x \right)+g\left( x \right) \right]dx}=10$ và $\int\limits_{0}^{2}{\left[ 3f\left( x \right)-g\left( x \right) \right]dx}=6.$ Tính $\int\limits_{2019}^{2021}{f\left( 2021-x \right)dx}+3\int\limits_{0}^{1}{g\left( 2x \right)dx}$ :
A. 7
B. 13
C. 5
D. 6
A. 7
B. 13
C. 5
D. 6
Phương pháp:
- Sử dụng tính chất tích phân: $\int\limits_{a}^{b}{\left[ f\left( x \right)+g\left( x \right) \right]dx}=\int\limits_{a}^{b}{f\left( x \right)dx}+\int\limits_{a}^{b}{g\left( x \right)dx},\int\limits_{a}^{b}{kf\left( x \right)dx}=k\int\limits_{a}^{b}{f\left( x \right)dx}\left( k\ne 0 \right)$ tìm $\int\limits_{0}^{2}{f\left( x \right)dx},\int\limits_{0}^{2}{g\left( x \right)dx}.$
- Sử dụng phương pháp đưa biến vào vi phân tính từng tích phân $\int\limits_{2019}^{2021}{f\left( 2021-x \right)dx},\int\limits_{0}^{1}{g\left( 2x \right)dx}.$
Cách giải:
Ta có: $\left\{ \begin{aligned}
& \int\limits_{0}^{2}{\left[ f\left( x \right)+g\left( x \right) \right]dx}=10 \\
& \int\limits_{0}^{2}{\left[ 3f\left( x \right)-g\left( x \right) \right]dx}=6 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& \int\limits_{0}^{2}{f\left( x \right)dx}+\int\limits_{0}^{2}{g\left( x \right)dx}=10 \\
& 3\int\limits_{0}^{2}{f\left( x \right)dx}-\int\limits_{0}^{2}{g\left( x \right)dx}=6 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& \int\limits_{0}^{2}{f\left( x \right)dx}=4 \\
& \int\limits_{0}^{2}{g\left( x \right)dx}=6 \\
\end{aligned} \right..$
Ta có:
$\int\limits_{2019}^{2021}{f\left( 2021-x \right)dx}=-\int\limits_{2019}^{2021}{f\left( 2021-x \right)d\left( 2021-x \right)dx}=-\int\limits_{2}^{0}{f\left( t \right)dt}=\int\limits_{0}^{2}{f\left( x \right)dx}=4$
$\int\limits_{0}^{1}{g\left( 2x \right)dx}=\dfrac{1}{2}\int\limits_{0}^{1}{g\left( 2x \right)d\left( 2x \right)}=\dfrac{1}{2}\int\limits_{0}^{2}{g\left( u \right)du}=\dfrac{1}{2}\int\limits_{0}^{2}{g\left( x \right)dx}=3$
Vậy $\int\limits_{2019}^{2021}{f\left( 2021-x \right)dx}+3\int\limits_{0}^{1}{g\left( 2x \right)dx}=4+3.3=13.$
- Sử dụng tính chất tích phân: $\int\limits_{a}^{b}{\left[ f\left( x \right)+g\left( x \right) \right]dx}=\int\limits_{a}^{b}{f\left( x \right)dx}+\int\limits_{a}^{b}{g\left( x \right)dx},\int\limits_{a}^{b}{kf\left( x \right)dx}=k\int\limits_{a}^{b}{f\left( x \right)dx}\left( k\ne 0 \right)$ tìm $\int\limits_{0}^{2}{f\left( x \right)dx},\int\limits_{0}^{2}{g\left( x \right)dx}.$
- Sử dụng phương pháp đưa biến vào vi phân tính từng tích phân $\int\limits_{2019}^{2021}{f\left( 2021-x \right)dx},\int\limits_{0}^{1}{g\left( 2x \right)dx}.$
Cách giải:
Ta có: $\left\{ \begin{aligned}
& \int\limits_{0}^{2}{\left[ f\left( x \right)+g\left( x \right) \right]dx}=10 \\
& \int\limits_{0}^{2}{\left[ 3f\left( x \right)-g\left( x \right) \right]dx}=6 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& \int\limits_{0}^{2}{f\left( x \right)dx}+\int\limits_{0}^{2}{g\left( x \right)dx}=10 \\
& 3\int\limits_{0}^{2}{f\left( x \right)dx}-\int\limits_{0}^{2}{g\left( x \right)dx}=6 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& \int\limits_{0}^{2}{f\left( x \right)dx}=4 \\
& \int\limits_{0}^{2}{g\left( x \right)dx}=6 \\
\end{aligned} \right..$
Ta có:
$\int\limits_{2019}^{2021}{f\left( 2021-x \right)dx}=-\int\limits_{2019}^{2021}{f\left( 2021-x \right)d\left( 2021-x \right)dx}=-\int\limits_{2}^{0}{f\left( t \right)dt}=\int\limits_{0}^{2}{f\left( x \right)dx}=4$
$\int\limits_{0}^{1}{g\left( 2x \right)dx}=\dfrac{1}{2}\int\limits_{0}^{1}{g\left( 2x \right)d\left( 2x \right)}=\dfrac{1}{2}\int\limits_{0}^{2}{g\left( u \right)du}=\dfrac{1}{2}\int\limits_{0}^{2}{g\left( x \right)dx}=3$
Vậy $\int\limits_{2019}^{2021}{f\left( 2021-x \right)dx}+3\int\limits_{0}^{1}{g\left( 2x \right)dx}=4+3.3=13.$
Đáp án B.