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Câu hỏi: Cho hàm số $f\left( x \right)$ có đạo hàm ${f}'\left( x \right)$ và thỏa mãn $A=\int\limits_{0}^{1}{\left( 2\text{x}+1 \right){f}'\left( x \right)d\text{x}}=10$, $3f\left( 1 \right)-f\left( 0 \right)=12$. Tính $I=\int\limits_{0}^{1}{f\left( x \right)d\text{x}}$.
A. $I=1$
B. $I=-2$
C. $I=2$
D. $I=-1$
A. $I=1$
B. $I=-2$
C. $I=2$
D. $I=-1$
Đặt $\left\{ \begin{aligned}
& u=2\text{x}+1 \\
& dv={f}'\left( x \right)d\text{x} \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=2\text{dx} \\
& v=f\left( x \right) \\
\end{aligned} \right.$.
$A=\int\limits_{0}^{1}{\left( 2\text{x}+1 \right){f}'\left( x \right)d\text{x}}$
$A=\left. \left( 2\text{x}+1 \right)f\left( x \right) \right|_{0}^{1}-2\int\limits_{0}^{1}{f\left( x \right)d\text{x}}$
$A=3f\left( 1 \right)-f\left( 0 \right)-2\int\limits_{0}^{1}{f\left( x \right)d\text{x}}$.
Mà $A=10;3f\left( 1 \right)-f\left( 0 \right)=12$. Từ đó suy ra: $12-2\int\limits_{0}^{1}{f\left( x \right)d\text{x}}=10\Leftrightarrow I=1$.
& u=2\text{x}+1 \\
& dv={f}'\left( x \right)d\text{x} \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=2\text{dx} \\
& v=f\left( x \right) \\
\end{aligned} \right.$.
$A=\int\limits_{0}^{1}{\left( 2\text{x}+1 \right){f}'\left( x \right)d\text{x}}$
$A=\left. \left( 2\text{x}+1 \right)f\left( x \right) \right|_{0}^{1}-2\int\limits_{0}^{1}{f\left( x \right)d\text{x}}$
$A=3f\left( 1 \right)-f\left( 0 \right)-2\int\limits_{0}^{1}{f\left( x \right)d\text{x}}$.
Mà $A=10;3f\left( 1 \right)-f\left( 0 \right)=12$. Từ đó suy ra: $12-2\int\limits_{0}^{1}{f\left( x \right)d\text{x}}=10\Leftrightarrow I=1$.
Đáp án A.