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Câu hỏi: Cho $f\left( x \right)$ là hàm số có đạo hàm liên tục trên đoạn $\left[ 0 ; 1 \right]$ và $f\left( 1 \right)=-\frac{1}{18}$, $\int\limits_{0}^{1}{x{f}'\left( x \right)}\text{d}x=\frac{1}{36}$. Giá trị của $\int\limits_{0}^{1}{f\left( x \right)}\text{d}x$ bằng
A. $-\frac{1}{12}$.
B. $\frac{1}{36}$.
C. $\frac{1}{12}$
D. $-\frac{1}{36}$.
A. $-\frac{1}{12}$.
B. $\frac{1}{36}$.
C. $\frac{1}{12}$
D. $-\frac{1}{36}$.
Ta có: $\int\limits_{0}^{1}{x{f}'\left( x \right)}\text{d}x=x.f\left( x \right)\left| \begin{aligned}
& 1 \\
& 0 \\
\end{aligned} \right.-\int\limits_{0}^{1}{f\left( x \right)}\text{d}x=f\left( 1 \right)-\int\limits_{0}^{1}{f\left( x \right)}\text{d}x$.
Ta có $\int\limits_{0}^{1}{x{f}'\left( x \right)}\text{d}x=\frac{1}{36}$, $f\left( 1 \right)=-\frac{1}{18}$, suy ra $\int\limits_{0}^{1}{f\left( x \right)}\text{d}x=-\frac{1}{18}-\frac{1}{36}=-\frac{1}{12}$.
& 1 \\
& 0 \\
\end{aligned} \right.-\int\limits_{0}^{1}{f\left( x \right)}\text{d}x=f\left( 1 \right)-\int\limits_{0}^{1}{f\left( x \right)}\text{d}x$.
Ta có $\int\limits_{0}^{1}{x{f}'\left( x \right)}\text{d}x=\frac{1}{36}$, $f\left( 1 \right)=-\frac{1}{18}$, suy ra $\int\limits_{0}^{1}{f\left( x \right)}\text{d}x=-\frac{1}{18}-\frac{1}{36}=-\frac{1}{12}$.
Đáp án A.