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Câu hỏi: Cho hai hàm số $f\left( x \right)=a{{x}^{3}}+bx+c$ ; $g\left( x \right)=b{{x}^{3}}+ax+c$, $\left( a>0 \right)$ có đồ thị như hình vẽ bên. Gọi ${{S}_{1}}$, ${{S}_{2}}$ là diện tích hình phẳng được gạch trong hình vẽ. Khi ${{S}_{1}}+{{S}_{2}}=3$ thì $\int\limits_{0}^{1}{f\left( x \right)dx}$ bằng
A. $3$.
B. $-3$.
C. $6$.
D. $-6$.
Phương trình hoành độ giao điểm $a{{x}^{3}}+bx+c=b{{x}^{3}}+ax+c$ $\Leftrightarrow \left( a-b \right){{x}^{3}}+\left( b-a \right)x=0$ $\Leftrightarrow \left( a-b \right)\left[ {{x}^{3}}-x \right]=0$ $\Leftrightarrow \left[ \begin{aligned}
& x=0 \\
& x=\pm 1 \\
\end{aligned} \right.$.
Cách 1:
Có $\left\{ \begin{aligned}
& {{S}_{1}}=\int\limits_{-1}^{0}{\left( f\left( x \right)-g\left( x \right) \right)dx}=\left( a-b \right)\int\limits_{-1}^{0}{\left( {{x}^{3}}-x \right)dx}=\dfrac{1}{4}\left( a-b \right) \\
& {{S}_{3}}=\int\limits_{0}^{1}{\left( g\left( x \right)-f\left( x \right) \right)dx}=\left( a-b \right)\int\limits_{0}^{1}{-\left( {{x}^{3}}-x \right)dx}=\dfrac{1}{4}\left( a-b \right) \\
\end{aligned} \right. $ $ \Rightarrow {{S}_{1}}={{S}_{3}}$.
Vậy ${{S}_{1}}+{{S}_{2}}=3$ $\Leftrightarrow {{S}_{3}}+{{S}_{2}}=3$ $\Leftrightarrow \int\limits_{0}^{1}{\left( g\left( x \right)-f\left( x \right) \right)dx}+\int\limits_{0}^{1}{-g\left( x \right)dx}=3$ $\Leftrightarrow \int\limits_{0}^{1}{f\left( x \right)dx}=-3$.
Cách 2:
${{S}_{1}}=\int\limits_{-1}^{0}{\left( f\left( x \right)-g\left( x \right) \right)dx}=\left( a-b \right)\int\limits_{-1}^{0}{\left( {{x}^{3}}-x \right)dx}=\dfrac{1}{4}\left( a-b \right)$ ;
${{S}_{2}}=-\int\limits_{0}^{1}{g\left( x \right)dx}=-\int\limits_{0}^{1}{\left( b{{x}^{3}}+ax+c \right)dx}=-\left( \dfrac{b}{4}+\dfrac{a}{2}+c \right)$.
Vậy ${{S}_{1}}+{{S}_{2}}=3$ $\Leftrightarrow \dfrac{1}{4}\left( a-b \right)-\dfrac{b}{4}-\dfrac{a}{2}-c=3$ $\Leftrightarrow a+2b+4c=-12$.
Suy ra $\int\limits_{0}^{1}{f\left( x \right)dx}=\int\limits_{0}^{1}{\left( a{{x}^{3}}+bx+c \right)dx}=\dfrac{a}{4}+\dfrac{b}{2}+c=\dfrac{a+2b+4c}{4}=-3$.
B. $-3$.
C. $6$.
D. $-6$.
Phương trình hoành độ giao điểm $a{{x}^{3}}+bx+c=b{{x}^{3}}+ax+c$ $\Leftrightarrow \left( a-b \right){{x}^{3}}+\left( b-a \right)x=0$ $\Leftrightarrow \left( a-b \right)\left[ {{x}^{3}}-x \right]=0$ $\Leftrightarrow \left[ \begin{aligned}
& x=0 \\
& x=\pm 1 \\
\end{aligned} \right.$.
Cách 1:
Có $\left\{ \begin{aligned}
& {{S}_{1}}=\int\limits_{-1}^{0}{\left( f\left( x \right)-g\left( x \right) \right)dx}=\left( a-b \right)\int\limits_{-1}^{0}{\left( {{x}^{3}}-x \right)dx}=\dfrac{1}{4}\left( a-b \right) \\
& {{S}_{3}}=\int\limits_{0}^{1}{\left( g\left( x \right)-f\left( x \right) \right)dx}=\left( a-b \right)\int\limits_{0}^{1}{-\left( {{x}^{3}}-x \right)dx}=\dfrac{1}{4}\left( a-b \right) \\
\end{aligned} \right. $ $ \Rightarrow {{S}_{1}}={{S}_{3}}$.
Cách 2:
${{S}_{1}}=\int\limits_{-1}^{0}{\left( f\left( x \right)-g\left( x \right) \right)dx}=\left( a-b \right)\int\limits_{-1}^{0}{\left( {{x}^{3}}-x \right)dx}=\dfrac{1}{4}\left( a-b \right)$ ;
${{S}_{2}}=-\int\limits_{0}^{1}{g\left( x \right)dx}=-\int\limits_{0}^{1}{\left( b{{x}^{3}}+ax+c \right)dx}=-\left( \dfrac{b}{4}+\dfrac{a}{2}+c \right)$.
Vậy ${{S}_{1}}+{{S}_{2}}=3$ $\Leftrightarrow \dfrac{1}{4}\left( a-b \right)-\dfrac{b}{4}-\dfrac{a}{2}-c=3$ $\Leftrightarrow a+2b+4c=-12$.
Suy ra $\int\limits_{0}^{1}{f\left( x \right)dx}=\int\limits_{0}^{1}{\left( a{{x}^{3}}+bx+c \right)dx}=\dfrac{a}{4}+\dfrac{b}{2}+c=\dfrac{a+2b+4c}{4}=-3$.
Đáp án B.