Google
Câu hỏi: Cho hàm số $f\left( x \right)$ có đạo hàm liên tục trên $\mathbb{R}$ thỏa mãn $2f\left( x \right)+xf'\left( x \right)=2x+1$ và $f\left( 1 \right)=-3.$ Khi đó $\int\limits_{0}^{1}{f\left( x \right)dx}$ bằng
A. $\dfrac{5}{2}$
B. 2
C. 5
D. $-1.$
A. $\dfrac{5}{2}$
B. 2
C. 5
D. $-1.$
Đặt $I=\int\limits_{0}^{1}{f\left( x \right)dx}$
Ta có $\int\limits_{0}^{1}{\left[ 2f\left( x \right)+xf'\left( x \right) \right]dx}=\int\limits_{0}^{1}{\left( 2x+1 \right)dx}=2.$
Mặt khác $\int\limits_{0}^{1}{\left[ 2f\left( x \right)+xf'\left( x \right) \right]dx}=2I+\int\limits_{0}^{1}{xf'\left( x \right)dx}=2I+\left( xf\left( x \right)\left| \begin{aligned}
& 1 \\
& 0 \\
\end{aligned} \right.-\int\limits_{0}^{1}{f\left( x \right)dx} \right)=2I+f\left( 1 \right)-I.$
Suy ra: $I-3=2\Rightarrow I=5.$
Ta có $\int\limits_{0}^{1}{\left[ 2f\left( x \right)+xf'\left( x \right) \right]dx}=\int\limits_{0}^{1}{\left( 2x+1 \right)dx}=2.$
Mặt khác $\int\limits_{0}^{1}{\left[ 2f\left( x \right)+xf'\left( x \right) \right]dx}=2I+\int\limits_{0}^{1}{xf'\left( x \right)dx}=2I+\left( xf\left( x \right)\left| \begin{aligned}
& 1 \\
& 0 \\
\end{aligned} \right.-\int\limits_{0}^{1}{f\left( x \right)dx} \right)=2I+f\left( 1 \right)-I.$
Suy ra: $I-3=2\Rightarrow I=5.$
Đáp án C.