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Câu hỏi: Cho hàm số $f\left( x \right)$ thỏa mãn $\int\limits_{0}^{1}{{{x}^{2}}.f'\left( x \right)dx}=-10$ và $f\left( 1 \right)=-2$. Tích phân $\int\limits_{0}^{1}{x.f\left( x \right)dx}$ bằng:
A. 4.
B. $-4.$
C. 6.
D. $-6.$
A. 4.
B. $-4.$
C. 6.
D. $-6.$
Ta có $\begin{aligned}
& -10=\int\limits_{0}^{1}{{{x}^{2}}.f'\left( x \right)dx}=\int\limits_{0}^{1}{{{x}^{2}}d\left[ f\left( x \right) \right]dx}={{x}^{2}}.f\left( x \right)\left| \begin{aligned}
& ^{1} \\
& _{0} \\
\end{aligned} \right.-\int\limits_{0}^{1}{f\left( x \right)d\left( {{x}^{2}} \right)} \\
& \ \ \ \ \ \ =f\left( 1 \right)-\int\limits_{0}^{1}{2x.f\left( x \right)dx}=-2-2\int\limits_{0}^{1}{x.f\left( x \right)dx}\Rightarrow \int\limits_{0}^{1}{x.f\left( x \right)dx}=4. \\
\end{aligned}$
& -10=\int\limits_{0}^{1}{{{x}^{2}}.f'\left( x \right)dx}=\int\limits_{0}^{1}{{{x}^{2}}d\left[ f\left( x \right) \right]dx}={{x}^{2}}.f\left( x \right)\left| \begin{aligned}
& ^{1} \\
& _{0} \\
\end{aligned} \right.-\int\limits_{0}^{1}{f\left( x \right)d\left( {{x}^{2}} \right)} \\
& \ \ \ \ \ \ =f\left( 1 \right)-\int\limits_{0}^{1}{2x.f\left( x \right)dx}=-2-2\int\limits_{0}^{1}{x.f\left( x \right)dx}\Rightarrow \int\limits_{0}^{1}{x.f\left( x \right)dx}=4. \\
\end{aligned}$
Đáp án A.