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Câu hỏi: Hình dưới đây vẽ đồ thị các hàm số $f\left( x \right)=-{{x}^{2}}-2x+1$ và $g\left( x \right)=-\dfrac{1}{2}{{x}^{3}}-\dfrac{5}{2}{{x}^{2}}-\dfrac{3}{2}x+\dfrac{5}{2}$.
Diện tích phần gạch chéo trong hình bằng
A. $\int\limits_{-3}^{-1}{\left[ f\left( x \right)-g\left( x \right) \right]} \text{d}x+\int\limits_{-1}^{1}{\left[ f\left( x \right)-g\left( x \right) \right]} \text{d}x$.
B. $\int\limits_{-3}^{-1}{\left[ g\left( x \right)-f\left( x \right) \right]} \text{d}x+\int\limits_{-1}^{1}{\left[ f\left( x \right)-g\left( x \right) \right]} \text{d}x$.
C. $\int\limits_{-3}^{-1}{\left[ g\left( x \right)-f\left( x \right) \right]} \text{d}x+\int\limits_{-1}^{1}{\left[ g\left( x \right)-f\left( x \right) \right]} \text{d}x$.
D. $\int\limits_{-3}^{-1}{\left[ f\left( x \right)-g\left( x \right) \right]} \text{d}x+\int\limits_{-1}^{1}{\left[ g\left( x \right)-f\left( x \right) \right]} \text{d}x$.
Diện tích phần gạch chéo trong hình bằng
A. $\int\limits_{-3}^{-1}{\left[ f\left( x \right)-g\left( x \right) \right]} \text{d}x+\int\limits_{-1}^{1}{\left[ f\left( x \right)-g\left( x \right) \right]} \text{d}x$.
B. $\int\limits_{-3}^{-1}{\left[ g\left( x \right)-f\left( x \right) \right]} \text{d}x+\int\limits_{-1}^{1}{\left[ f\left( x \right)-g\left( x \right) \right]} \text{d}x$.
C. $\int\limits_{-3}^{-1}{\left[ g\left( x \right)-f\left( x \right) \right]} \text{d}x+\int\limits_{-1}^{1}{\left[ g\left( x \right)-f\left( x \right) \right]} \text{d}x$.
D. $\int\limits_{-3}^{-1}{\left[ f\left( x \right)-g\left( x \right) \right]} \text{d}x+\int\limits_{-1}^{1}{\left[ g\left( x \right)-f\left( x \right) \right]} \text{d}x$.
Dựa vào đồ thị các hàm số $f\left( x \right)=-{{x}^{2}}-2x+1$ và $g\left( x \right)=-\dfrac{1}{2}{{x}^{3}}-\dfrac{5}{2}{{x}^{2}}-\dfrac{3}{2}x+\dfrac{5}{2}$ trên mặt phẳng tọa độ, ta thấy $\forall x\in \left[ -3;-1 \right]$ : $f\left( x \right)\ge g\left( x \right)$.
$\forall x\in \left[ -1;1 \right]$ : $g\left( x \right)\ge f\left( x \right)$.
Vậy diện tích $S$ phần gạch chéo trong hình bằng
$S=\int\limits_{-3}^{1}{\left| f\left( x \right)-g\left( x \right) \right|} \text{d}x=\int\limits_{-3}^{-1}{\left| f\left( x \right)-g\left( x \right) \right|} \text{d}x+\int\limits_{-1}^{1}{\left| f\left( x \right)-g\left( x \right) \right|} \text{d}x$
$=\int\limits_{-3}^{-1}{\left[ f\left( x \right)-g\left( x \right) \right]} \text{d}x+\int\limits_{-1}^{1}{\left[ g\left( x \right)-f\left( x \right) \right]} \text{d}x$.
$\forall x\in \left[ -1;1 \right]$ : $g\left( x \right)\ge f\left( x \right)$.
Vậy diện tích $S$ phần gạch chéo trong hình bằng
$S=\int\limits_{-3}^{1}{\left| f\left( x \right)-g\left( x \right) \right|} \text{d}x=\int\limits_{-3}^{-1}{\left| f\left( x \right)-g\left( x \right) \right|} \text{d}x+\int\limits_{-1}^{1}{\left| f\left( x \right)-g\left( x \right) \right|} \text{d}x$
$=\int\limits_{-3}^{-1}{\left[ f\left( x \right)-g\left( x \right) \right]} \text{d}x+\int\limits_{-1}^{1}{\left[ g\left( x \right)-f\left( x \right) \right]} \text{d}x$.
Đáp án D.