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