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Câu hỏi: Cho tích phân $\int\limits_{1}^{2}{\left[ 4f\left( x \right)-2x \right]\text{d}x}= 1$ khi đó $ \int\limits_{1}^{2}{f\left( x \right)\text{d}x}$ bằng
A. $3\cdot $
B. $-1.$
C. $1\cdot $
D. $-3\cdot $
A. $3\cdot $
B. $-1.$
C. $1\cdot $
D. $-3\cdot $
Ta có $\int\limits_{1}^{2}{\left[ 4f\left( x \right)-2x \right]\text{d}x}=4\int\limits_{1}^{2}{f\left( x \right)\text{d}x-\int\limits_{1}^{2}{2x} \text{d}x=}4\int\limits_{1}^{2}{f\left( x \right)} \text{d}x-3 =1$ $\Leftrightarrow 4 \int\limits_{1}^{2}{f\left( x \right)\text{d}x}=4$.
Vậy $ \int\limits_{1}^{2}{f\left( x \right)\text{d}x}=1$.
Vậy $ \int\limits_{1}^{2}{f\left( x \right)\text{d}x}=1$.
Đáp án C.