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Câu hỏi: Cho hàm số $f\left( x \right)$ liên tục trên đoạn $\left[ 0;10 \right]$ thỏa mãn $\int\limits_{0}^{10}{f\left( x \right)\text{d}x=7,}\int\limits_{2}^{10}{f\left( x \right)\text{d}x=1}$. Tính $P=\int\limits_{0}^{1}{f\left( 2x \right)\text{d}x}$.
A. $P=6$.
B. $P=-6$.
C. $P=3$.
D. $P=12$.
A. $P=6$.
B. $P=-6$.
C. $P=3$.
D. $P=12$.
Ta có: $\int\limits_{0}^{2}{f\left( x \right)\text{d}x=}\int\limits_{0}^{10}{f\left( x \right)\text{d}x-}\int\limits_{2}^{10}{f\left( x \right)\text{d}x=6}$.
Xét $P=\int\limits_{0}^{1}{f\left( 2x \right)\text{d}x}$. Đặt $t=2x\Rightarrow \text{d}t=2\text{d}x\Rightarrow \text{d}x=\dfrac{1}{2}\text{d}t$.
Đổi cận:
Lúc đó: $P=\int\limits_{0}^{1}{f\left( 2x \right)\text{d}x}=\dfrac{1}{2}\int\limits_{0}^{2}{f\left( t \right)\text{d}t}=\dfrac{1}{2}\int\limits_{0}^{2}{f\left( x \right)\text{d}}x=3$.
Xét $P=\int\limits_{0}^{1}{f\left( 2x \right)\text{d}x}$. Đặt $t=2x\Rightarrow \text{d}t=2\text{d}x\Rightarrow \text{d}x=\dfrac{1}{2}\text{d}t$.
Đổi cận:
Lúc đó: $P=\int\limits_{0}^{1}{f\left( 2x \right)\text{d}x}=\dfrac{1}{2}\int\limits_{0}^{2}{f\left( t \right)\text{d}t}=\dfrac{1}{2}\int\limits_{0}^{2}{f\left( x \right)\text{d}}x=3$.
Đáp án C.