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Câu hỏi: Cho $\int\limits_{0}^{4}{f\left( x \right)dx=3}$ và $\int\limits_{0}^{2}{g\left( 2x \right)dx=4}$. Tính $\int\limits_{0}^{4}{\left[ f\left( x \right)-g\left( x \right) \right]dx}$.
A. $\int\limits_{0}^{4}{\left[ f\left( x \right)-g\left( x \right) \right]dx}=1$.
B. $\int\limits_{0}^{4}{\left[ f\left( x \right)-g\left( x \right) \right]dx}=-1$.
C. $\int\limits_{0}^{4}{\left[ f\left( x \right)-g\left( x \right) \right]dx}=-5$.
D. $\int\limits_{0}^{4}{\left[ f\left( x \right)-g\left( x \right) \right]dx}=5$.
A. $\int\limits_{0}^{4}{\left[ f\left( x \right)-g\left( x \right) \right]dx}=1$.
B. $\int\limits_{0}^{4}{\left[ f\left( x \right)-g\left( x \right) \right]dx}=-1$.
C. $\int\limits_{0}^{4}{\left[ f\left( x \right)-g\left( x \right) \right]dx}=-5$.
D. $\int\limits_{0}^{4}{\left[ f\left( x \right)-g\left( x \right) \right]dx}=5$.
Ta có $\int\limits_{0}^{2}{g\left( 2x \right)dx}=\dfrac{1}{2}\int\limits_{0}^{2}{2g\left( 2x \right)dx}=\dfrac{1}{2}\int\limits_{0}^{2}{g\left( 2x \right)d\left( 2x \right)}=\dfrac{1}{2}\int\limits_{0}^{4}{g\left( t \right)dt}=\dfrac{1}{2}\int\limits_{0}^{4}{g\left( x \right)dx}$.
Suy ra $\int\limits_{0}^{4}{g\left( x \right)dx}=8$.
Vậy $\int\limits_{0}^{4}{\left[ f\left( x \right)-g\left( x \right) \right]dx}=-5$.
Suy ra $\int\limits_{0}^{4}{g\left( x \right)dx}=8$.
Vậy $\int\limits_{0}^{4}{\left[ f\left( x \right)-g\left( x \right) \right]dx}=-5$.
Đáp án C.