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Câu hỏi: Cho hàm số bậc ba $y=f(x).$ Đường thẳng $y=ax+b$ tạo với đường $y=f(x)$ hai miền phẳng có diện tích là ${{S}_{1}},{{S}_{2}}$ (hình vẽ bên).
Biết ${{S}_{1}}=\dfrac{5}{12}$ và $\int\limits_{0}^{1}{\left( 1-2x \right){f}'\left( 3x \right)\text{d}x}=-\dfrac{1}{2}$, giá trị của ${{S}_{2}}$ bằng
A. $\dfrac{8}{3}$.
B. $\dfrac{19}{4}$.
C. $\dfrac{13}{3}$.
D. $\dfrac{13}{6}$.
A. $\dfrac{8}{3}$.
B. $\dfrac{19}{4}$.
C. $\dfrac{13}{3}$.
D. $\dfrac{13}{6}$.
$\int\limits_{0}^{1}{\left( 1-2x \right){f}'\left( 3x \right)\text{d}x}=\int\limits_{0}^{1}{\left( 1-2x \right)\text{d}\left[ \dfrac{1}{3}f\left( 3x \right) \right]}=\dfrac{1}{3}\left. f\left( 3x \right)\left( 1-2x \right) \right|_{0}^{1}+\dfrac{2}{3}\int\limits_{0}^{1}{f\left( 3x \right)\text{d}x}$
$=\dfrac{-1}{3}f\left( 3 \right)-\dfrac{1}{3}f\left( 0 \right)+\dfrac{2}{9}\int\limits_{0}^{3}{f\left( x \right)\text{d}x}=\dfrac{2}{3}+\dfrac{2}{9}\int\limits_{0}^{3}{f\left( x \right)\text{d}x}=\dfrac{-1}{2}\Leftrightarrow \int\limits_{0}^{3}{f\left( x \right)\text{d}x}=\dfrac{-21}{4}$.
Khi đó ${{S}_{2}}=\left| \int\limits_{0}^{3}{f\left( x \right)\text{d}x} \right|-\left( {{S}_{OAB}}-{{S}_{1}} \right)=\dfrac{8}{3}$ với $A\left( 0;-2 \right)$, $B\left( 3;0 \right)$.
$=\dfrac{-1}{3}f\left( 3 \right)-\dfrac{1}{3}f\left( 0 \right)+\dfrac{2}{9}\int\limits_{0}^{3}{f\left( x \right)\text{d}x}=\dfrac{2}{3}+\dfrac{2}{9}\int\limits_{0}^{3}{f\left( x \right)\text{d}x}=\dfrac{-1}{2}\Leftrightarrow \int\limits_{0}^{3}{f\left( x \right)\text{d}x}=\dfrac{-21}{4}$.
Khi đó ${{S}_{2}}=\left| \int\limits_{0}^{3}{f\left( x \right)\text{d}x} \right|-\left( {{S}_{OAB}}-{{S}_{1}} \right)=\dfrac{8}{3}$ với $A\left( 0;-2 \right)$, $B\left( 3;0 \right)$.
Đáp án A.