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Câu hỏi: Cho hàm số $y=f\left( x \right)$ có đạo hàm ${f}'\left( x \right)$ liên tục trên $\mathbb{R}$, đồ thị hàm số $y={f}'\left( x \right)$ như hình vẽ. Biết $\int\limits_{0}^{3}{\left( x+1 \right){f}'\left( x \right)d\text{x}}=a$ và $\int\limits_{0}^{1}{\left| {f}'\left( x \right) \right|d\text{x}}=b$, $\int\limits_{1}^{3}{\left| {f}'\left( x \right) \right|d\text{x}}=c$, $f\left( 1 \right)=d$. Tích phân $\int\limits_{0}^{3}{f\left( x \right)d\text{x}}$ bằng
A. $-a+b-3c+2\text{d}$
B. $-a+b-4c+3\text{d}$
C. $-a+b+4c-5\text{d}$
D. $-a-b-4c+5\text{d}$
A. $-a+b-3c+2\text{d}$
B. $-a+b-4c+3\text{d}$
C. $-a+b+4c-5\text{d}$
D. $-a-b-4c+5\text{d}$
Đặt $\left\{ \begin{aligned}
& u=x+1 \\
& dv={f}'\left( x \right)d\text{x} \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=d\text{x} \\
& v=f\left( x \right) \\
\end{aligned} \right.$ ta có:
$\int\limits_{0}^{3}{\left( x+1 \right){f}'\left( x \right)d\text{x}}=\left. \left( x+1 \right)f\left( x \right) \right|_{0}^{3}-\int\limits_{0}^{3}{f\left( x \right)d\text{x}}=4f\left( 3 \right)-f\left( 0 \right)-\int\limits_{0}^{3}{f\left( x \right)d\text{x}}=a$.
Mặt khác $\int\limits_{0}^{1}{\left| {f}'\left( x \right) \right|\text{d}x}=\int\limits_{0}^{1}{{f}'\left( x \right)dx}=f\left( 1 \right)-f\left( 0 \right)=b\Leftrightarrow f\left( 0 \right)=d-b$.
Lại có $\int\limits_{1}^{3}{\left| {f}'\left( x \right) \right|\text{d}x}=-\int\limits_{1}^{3}{{f}'\left( x \right)dx}=f\left( 1 \right)-f\left( 3 \right)=d-f\left( 3 \right)=c\Rightarrow f\left( 3 \right)=d-c$.
Thế vào ta được $4\left( d-c \right)-\left( d-b \right)-I=a\Leftrightarrow 3d+b-4c-a=I.$
& u=x+1 \\
& dv={f}'\left( x \right)d\text{x} \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=d\text{x} \\
& v=f\left( x \right) \\
\end{aligned} \right.$ ta có:
$\int\limits_{0}^{3}{\left( x+1 \right){f}'\left( x \right)d\text{x}}=\left. \left( x+1 \right)f\left( x \right) \right|_{0}^{3}-\int\limits_{0}^{3}{f\left( x \right)d\text{x}}=4f\left( 3 \right)-f\left( 0 \right)-\int\limits_{0}^{3}{f\left( x \right)d\text{x}}=a$.
Mặt khác $\int\limits_{0}^{1}{\left| {f}'\left( x \right) \right|\text{d}x}=\int\limits_{0}^{1}{{f}'\left( x \right)dx}=f\left( 1 \right)-f\left( 0 \right)=b\Leftrightarrow f\left( 0 \right)=d-b$.
Lại có $\int\limits_{1}^{3}{\left| {f}'\left( x \right) \right|\text{d}x}=-\int\limits_{1}^{3}{{f}'\left( x \right)dx}=f\left( 1 \right)-f\left( 3 \right)=d-f\left( 3 \right)=c\Rightarrow f\left( 3 \right)=d-c$.
Thế vào ta được $4\left( d-c \right)-\left( d-b \right)-I=a\Leftrightarrow 3d+b-4c-a=I.$
Đáp án B.