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Câu hỏi: Cho hàm số $f\left( x \right)$ có đạo hàm $f'\left( x \right)$ liên tục trên $\mathbb{R}$ và có đồ thị của hàm số $f'\left( x \right)$ như hình.
Biết $\int\limits_{0}^{3}{\left( x+1 \right)}f'\left( x \right)dx=a$ và $\int\limits_{0}^{1}{\left| f'\left( x \right) \right|dx}=b,\int\limits_{1}^{3}{\left| f'\left( x \right) \right|dx}=c,f\left( 1 \right)=d$. Tích phân $\int\limits_{0}^{3}{f\left( x \right)dx}$ bằng
A. $-a+b+4c-5d.$
B. $-a+b-3c+2d.$
C. $-a+b-4c+3d.$
D. $-a-b-4c+5d.$
Biết $\int\limits_{0}^{3}{\left( x+1 \right)}f'\left( x \right)dx=a$ và $\int\limits_{0}^{1}{\left| f'\left( x \right) \right|dx}=b,\int\limits_{1}^{3}{\left| f'\left( x \right) \right|dx}=c,f\left( 1 \right)=d$. Tích phân $\int\limits_{0}^{3}{f\left( x \right)dx}$ bằng
B. $-a+b-3c+2d.$
C. $-a+b-4c+3d.$
D. $-a-b-4c+5d.$
Tích phân từng phần có: $\int\limits_{0}^{3}{\left( x+1 \right)f'\left( x \right)}dx=\int\limits_{0}^{3}{\left( x+1 \right)d\left( f\left( x \right) \right)}$
$=\left( x+1 \right)\left. f\left( x \right) \right|_{0}^{3}-\int\limits_{0}^{3}{f\left( x \right)}dx=4f\left( 3 \right)-f\left( 0 \right)-\int\limits_{0}^{3}{f\left( x \right)}dx$
Suy ra $\int\limits_{0}^{3}{f\left( x \right)}dx=4f\left( 3 \right)-f\left( 0 \right)-\int\limits_{0}^{3}{\left( x+1 \right)f'\left( x \right)}dx=4f\left( 3 \right)-f\left( 0 \right)-a (1)$
$b=\int\limits_{0}^{1}{\left| f'\left( x \right) \right|}dx=\int\limits_{0}^{1}{f'\left( x \right)}dx=f\left( 1 \right)-f\left( 0 \right)=d-f\left( 0 \right)\Rightarrow f\left( 0 \right)=b-d (2)$
$c=\int\limits_{1}^{3}{\left| f'\left( x \right) \right|}dx=-\int\limits_{1}^{3}{f'\left( x \right)}dx=f\left( 1 \right)-f\left( 3 \right)=d-f\left( 3 \right)\Rightarrow f\left( 3 \right)=d-c \left( 3 \right)$
Từ $(1),(2),(3)\Rightarrow \int\limits_{0}^{3}{f\left( x \right)}dx=4\left( d-c \right)-\left( d-b \right)-a=-a+b-4c+3d$
$=\left( x+1 \right)\left. f\left( x \right) \right|_{0}^{3}-\int\limits_{0}^{3}{f\left( x \right)}dx=4f\left( 3 \right)-f\left( 0 \right)-\int\limits_{0}^{3}{f\left( x \right)}dx$
Suy ra $\int\limits_{0}^{3}{f\left( x \right)}dx=4f\left( 3 \right)-f\left( 0 \right)-\int\limits_{0}^{3}{\left( x+1 \right)f'\left( x \right)}dx=4f\left( 3 \right)-f\left( 0 \right)-a (1)$
$b=\int\limits_{0}^{1}{\left| f'\left( x \right) \right|}dx=\int\limits_{0}^{1}{f'\left( x \right)}dx=f\left( 1 \right)-f\left( 0 \right)=d-f\left( 0 \right)\Rightarrow f\left( 0 \right)=b-d (2)$
$c=\int\limits_{1}^{3}{\left| f'\left( x \right) \right|}dx=-\int\limits_{1}^{3}{f'\left( x \right)}dx=f\left( 1 \right)-f\left( 3 \right)=d-f\left( 3 \right)\Rightarrow f\left( 3 \right)=d-c \left( 3 \right)$
Từ $(1),(2),(3)\Rightarrow \int\limits_{0}^{3}{f\left( x \right)}dx=4\left( d-c \right)-\left( d-b \right)-a=-a+b-4c+3d$
Đáp án C.