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Câu hỏi: Cho hàm số $y=f\left( x \right)$ có đạo hàm trên $\mathbb{R}$, đồ thị hàm số $y=f\left( x \right)$ như hình vẽ. Biết diện tích hình phẳng phần sọc kẻ bằng $3$. Tính giá trị của biểu thức:
$T=\int\limits_{1}^{2}{{f}'\left( x+1 \right)\text{dx}}+\int\limits_{2}^{3}{{f}'\left( x-1 \right)\text{dx}}+\int\limits_{3}^{4}{f\left( 2x-8 \right)\text{dx}}$
A. $T=\dfrac{9}{2}$.
B. $T=6$.
C. $T=0$.
D. $T=\dfrac{3}{2}$.
$T=\int\limits_{1}^{2}{{f}'\left( x+1 \right)\text{dx}}+\int\limits_{2}^{3}{{f}'\left( x-1 \right)\text{dx}}+\int\limits_{3}^{4}{f\left( 2x-8 \right)\text{dx}}$
A. $T=\dfrac{9}{2}$.
B. $T=6$.
C. $T=0$.
D. $T=\dfrac{3}{2}$.
Diện tích phần kẻ sọc là: $S=\int\limits_{-2}^{0}{\left| f\left( x \right) \right|\text{dx}}$ $=3$.
Vì $f\left( x \right)\le 0$ $\forall x\in \left[ -2;0 \right]$ $\Rightarrow 3=\int\limits_{-2}^{0}{\left| f\left( x \right) \right|\text{dx}}=\int\limits_{-2}^{0}{\left[ -f\left( x \right) \right]\text{dx}}$ $\Leftrightarrow \int\limits_{-2}^{0}{f\left( x \right)\text{dx}}=-3$.
Tính $I=\int\limits_{3}^{4}{f\left( 2x-8 \right)\text{dx}}$.
Đặt $t=2x-8$ $\Rightarrow \text{dt}=2\text{dx}$ ; $x=3\Rightarrow t=-2$ ; $x=4\Rightarrow t=0$.
Suy ra: $I=\int\limits_{-2}^{0}{f\left( t \right)\text{.}\dfrac{1}{2}\text{dt}}$ $=\dfrac{1}{2}\int\limits_{-2}^{0}{f\left( x \right)\text{dx}}$ $=-\dfrac{3}{2}$.
Vậy $T=\int\limits_{1}^{2}{{f}'\left( x+1 \right)\text{dx}}+\int\limits_{2}^{3}{{f}'\left( x-1 \right)\text{dx}}+\int\limits_{3}^{4}{f\left( 2x-8 \right)\text{dx}}$
$=\left. f\left( x+1 \right) \right|_{1}^{2}+\left. f\left( x-1 \right) \right|_{2}^{3}+I$ $=f\left( 3 \right)-f\left( 2 \right)+f\left( 2 \right)-f\left( 1 \right)-\dfrac{3}{2}$ $=2-\left( -1 \right)-\dfrac{3}{2}=\dfrac{3}{2}$.
Vì $f\left( x \right)\le 0$ $\forall x\in \left[ -2;0 \right]$ $\Rightarrow 3=\int\limits_{-2}^{0}{\left| f\left( x \right) \right|\text{dx}}=\int\limits_{-2}^{0}{\left[ -f\left( x \right) \right]\text{dx}}$ $\Leftrightarrow \int\limits_{-2}^{0}{f\left( x \right)\text{dx}}=-3$.
Tính $I=\int\limits_{3}^{4}{f\left( 2x-8 \right)\text{dx}}$.
Đặt $t=2x-8$ $\Rightarrow \text{dt}=2\text{dx}$ ; $x=3\Rightarrow t=-2$ ; $x=4\Rightarrow t=0$.
Suy ra: $I=\int\limits_{-2}^{0}{f\left( t \right)\text{.}\dfrac{1}{2}\text{dt}}$ $=\dfrac{1}{2}\int\limits_{-2}^{0}{f\left( x \right)\text{dx}}$ $=-\dfrac{3}{2}$.
Vậy $T=\int\limits_{1}^{2}{{f}'\left( x+1 \right)\text{dx}}+\int\limits_{2}^{3}{{f}'\left( x-1 \right)\text{dx}}+\int\limits_{3}^{4}{f\left( 2x-8 \right)\text{dx}}$
$=\left. f\left( x+1 \right) \right|_{1}^{2}+\left. f\left( x-1 \right) \right|_{2}^{3}+I$ $=f\left( 3 \right)-f\left( 2 \right)+f\left( 2 \right)-f\left( 1 \right)-\dfrac{3}{2}$ $=2-\left( -1 \right)-\dfrac{3}{2}=\dfrac{3}{2}$.
Đáp án D.