Google
Câu hỏi: : Cho hàm số $f\left( x \right)$ có đạo hàm cấp 2 liên tục trên $\mathbb{R}$ thỏa mãn số nguyên $x$ thỏa mãn $f'\left( 1 \right)=2021,f\left( 1-x \right)+{{x}^{2}}f''\left( x \right)=3x,\forall x\in \mathbb{R}.$ Tính $I=\int\limits_{0}^{1}{xf'\left( x \right)dx}$
A. 674.
B. 673.
C. $\dfrac{2021}{3}.$
D. $\dfrac{2020}{3}.$
A. 674.
B. 673.
C. $\dfrac{2021}{3}.$
D. $\dfrac{2020}{3}.$
Ta có $f\left( 1-x \right)+{{x}^{2}}f''\left( x \right)=2x,\forall x\in \mathbb{R}\Rightarrow f\left( 1 \right)=0.$ Ta có
$\int\limits_{0}^{1}{\left( f\left( 1-x \right)+{{x}^{2}}f''\left( x \right) \right)dx}=\int\limits_{0}^{1}{2xdx}=1\Rightarrow 1=\int\limits_{0}^{1}{\left( f\left( x \right)+{{x}^{2}}f''\left( x \right) \right)dx}$ (Do $\int\limits_{0}^{1}{f\left( x \right)dx}=\int\limits_{0}^{1}{f\left( 1-x \right)dx}$ ).
Ta có:
$I=\int\limits_{0}^{1}{f\left( x \right)dx}+\int\limits_{0}^{1}{{{x}^{2}}f''\left( x \right)dx}=xf\left( x \right)\left| \begin{aligned}
& 1 \\
& 0 \\
\end{aligned} \right.-I+{{x}^{2}}f'\left( x \right)\left| \begin{aligned}
& 1 \\
& 0 \\
\end{aligned} \right.-2I=2021-3I\Rightarrow I=\dfrac{2020}{3}.$
$\int\limits_{0}^{1}{\left( f\left( 1-x \right)+{{x}^{2}}f''\left( x \right) \right)dx}=\int\limits_{0}^{1}{2xdx}=1\Rightarrow 1=\int\limits_{0}^{1}{\left( f\left( x \right)+{{x}^{2}}f''\left( x \right) \right)dx}$ (Do $\int\limits_{0}^{1}{f\left( x \right)dx}=\int\limits_{0}^{1}{f\left( 1-x \right)dx}$ ).
Ta có:
$I=\int\limits_{0}^{1}{f\left( x \right)dx}+\int\limits_{0}^{1}{{{x}^{2}}f''\left( x \right)dx}=xf\left( x \right)\left| \begin{aligned}
& 1 \\
& 0 \\
\end{aligned} \right.-I+{{x}^{2}}f'\left( x \right)\left| \begin{aligned}
& 1 \\
& 0 \\
\end{aligned} \right.-2I=2021-3I\Rightarrow I=\dfrac{2020}{3}.$
Đáp án D.