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Câu hỏi: Cho hàm số $f\left( x \right)$ có đạo hàm liên tục trên khoảng $\left( 0;+\infty \right)$ thỏa mãn ${f}'\left( x \right).f\left( x \right)=1,\forall x>0$.
Biết $f\left( 2 \right)=\left( 4 \right)=b,J=\int\limits_{1}^{2}{\dfrac{x}{f\left( 2x \right)}dx.}$ Tính $I=\int\limits_{1}^{3}{f\left( x+1 \right)dx}$ theo $a,b,c$.
A. $c-4b+2a.$
B. $4b-4c-2a.$
C. $4c-4b+2a.$
D. $2c-4b+2a.$
Biết $f\left( 2 \right)=\left( 4 \right)=b,J=\int\limits_{1}^{2}{\dfrac{x}{f\left( 2x \right)}dx.}$ Tính $I=\int\limits_{1}^{3}{f\left( x+1 \right)dx}$ theo $a,b,c$.
A. $c-4b+2a.$
B. $4b-4c-2a.$
C. $4c-4b+2a.$
D. $2c-4b+2a.$
Tính $I=\int\limits_{1}^{3}{f\left( x+1 \right)dx.}$ Đặt $t=x+1\Rightarrow I=\int\limits_{2}^{4}{f\left( t \right)}.dt$
Tính $J=\int\limits_{1}^{2}{\dfrac{x}{f\left( 2x \right)}dx=c.}$
Đặt $t=2x\Rightarrow J=\dfrac{1}{4}\int\limits_{2}^{4}{\dfrac{t}{f\left( t \right)}dt}=\dfrac{1}{4}\int\limits_{2}^{{}}{t{f}'\left( t \right)dt}=\dfrac{1}{4}\left( tf\left( t \right)\left| _{2}^{4} \right.-\int\limits_{2}^{4}{f\left( t \right)dt} \right)=c$
$=\dfrac{1}{4}\left( 4f\left( 4 \right)-2f\left( 2 \right)-\int\limits_{2}^{4}{f\left( t \right)dt} \right)=b-\dfrac{1}{2}a-\dfrac{1}{4}I\Rightarrow I=4b-4c-2a.$
Tính $J=\int\limits_{1}^{2}{\dfrac{x}{f\left( 2x \right)}dx=c.}$
Đặt $t=2x\Rightarrow J=\dfrac{1}{4}\int\limits_{2}^{4}{\dfrac{t}{f\left( t \right)}dt}=\dfrac{1}{4}\int\limits_{2}^{{}}{t{f}'\left( t \right)dt}=\dfrac{1}{4}\left( tf\left( t \right)\left| _{2}^{4} \right.-\int\limits_{2}^{4}{f\left( t \right)dt} \right)=c$
$=\dfrac{1}{4}\left( 4f\left( 4 \right)-2f\left( 2 \right)-\int\limits_{2}^{4}{f\left( t \right)dt} \right)=b-\dfrac{1}{2}a-\dfrac{1}{4}I\Rightarrow I=4b-4c-2a.$
Đáp án B.