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Câu hỏi: Cho $F\left( x \right)$ là một nguyên hàm của hàm $f\left( x \right)$ trên đoạn $\left[ -1 ;3 \right]$. Biết $F\left( -1 \right)=2$, $F\left( 3 \right)=\frac{11}{2}$. Tính tích phân $I=\int\limits_{-1}^{3}{\left[ 2f\left( x \right)-x \right]\text{d}x}$.
A. $I=\frac{7}{2}$.
B. $I=3$.
C. $I=11$.
D. $I=19$.
A. $I=\frac{7}{2}$.
B. $I=3$.
C. $I=11$.
D. $I=19$.
Ta có:
$I=\int\limits_{-1}^{3}{\left[ 2f\left( x \right)-x \right]\text{d}x}$ $=\int\limits_{-1}^{3}{2f\left( x \right)\text{d}x}-\int\limits_{-1}^{3}{x\text{d}x}$ $=2F\left( x \right)\left| \begin{aligned}
& 3 \\
& -1 \\
\end{aligned} \right.-\frac{{{x}^{2}}}{2}\left| \begin{aligned}
& 3 \\
& -1 \\
\end{aligned} \right.$$=2\left[ F\left( 3 \right)-F\left( -1 \right) \right]-\left( \frac{9}{2}-\frac{1}{2} \right) $ $ =2\left( \frac{11}{2}-2 \right)-\left( \frac{9}{2}-\frac{1}{2} \right)=3$.
$I=\int\limits_{-1}^{3}{\left[ 2f\left( x \right)-x \right]\text{d}x}$ $=\int\limits_{-1}^{3}{2f\left( x \right)\text{d}x}-\int\limits_{-1}^{3}{x\text{d}x}$ $=2F\left( x \right)\left| \begin{aligned}
& 3 \\
& -1 \\
\end{aligned} \right.-\frac{{{x}^{2}}}{2}\left| \begin{aligned}
& 3 \\
& -1 \\
\end{aligned} \right.$$=2\left[ F\left( 3 \right)-F\left( -1 \right) \right]-\left( \frac{9}{2}-\frac{1}{2} \right) $ $ =2\left( \frac{11}{2}-2 \right)-\left( \frac{9}{2}-\frac{1}{2} \right)=3$.
Đáp án B.