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Câu hỏi: Cho hàm số $f\left( x \right)$ liên tục trên đoạn $\left[ 0 ;1 \right]$, có đạo hàm ${f}'\left( x \right)$ thỏa mãn $\int\limits_{0}^{1}{\left( 2x+1 \right){f}'\left( x \right)\text{d}x}=10$ và $f\left( 0 \right)=3f\left( 1 \right)$. Tính $I=\int\limits_{0}^{1}{f\left( x \right)\text{d}x}$.
A. $I=-5$.
B. $I=-2$.
C. $I=2$.
D. $I=5$.
Ta có: $\int\limits_{0}^{1}{\left( 2x+1 \right){f}'\left( x \right)\text{d}x}=10$ $\Leftrightarrow \left( 2x+1 \right)f\left( x \right)\left| \begin{aligned}
& 1 \\
& 0 \\
\end{aligned} \right.-2\int\limits_{0}^{1}{f\left( x \right)\text{d}x}=10$
$\Leftrightarrow 3f\left( 1 \right)-f\left( 0 \right)-2\int\limits_{0}^{1}{f\left( x \right)\text{d}x}=10$ $\Leftrightarrow 0-2\int\limits_{0}^{1}{f\left( x \right)\text{d}x}=10$ $\Leftrightarrow \int\limits_{0}^{1}{f\left( x \right)\text{d}x}=-5$.
A. $I=-5$.
B. $I=-2$.
C. $I=2$.
D. $I=5$.
Đặt: $u=2x+1\Leftrightarrow \text{d}u=2\text{d}x$, $\text{d}v={f}'\left( x \right)\text{d}x$ chọn $v=f\left( x \right)$.Ta có: $\int\limits_{0}^{1}{\left( 2x+1 \right){f}'\left( x \right)\text{d}x}=10$ $\Leftrightarrow \left( 2x+1 \right)f\left( x \right)\left| \begin{aligned}
& 1 \\
& 0 \\
\end{aligned} \right.-2\int\limits_{0}^{1}{f\left( x \right)\text{d}x}=10$
$\Leftrightarrow 3f\left( 1 \right)-f\left( 0 \right)-2\int\limits_{0}^{1}{f\left( x \right)\text{d}x}=10$ $\Leftrightarrow 0-2\int\limits_{0}^{1}{f\left( x \right)\text{d}x}=10$ $\Leftrightarrow \int\limits_{0}^{1}{f\left( x \right)\text{d}x}=-5$.
Đáp án A.