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Câu hỏi: Diện tích hình phẳng giới hạn bởi $y={{x}^{3}},y=4x$ là:
A. 9.
B. 8.
C. 13.
D. 12.
A. 9.
B. 8.
C. 13.
D. 12.
Xét phương trình ${{x}^{3}}=4x\Leftrightarrow x\left( {{x}^{2}}-4 \right)=0\Leftrightarrow \left[ \begin{aligned}
& x=0 \\
& x=2 \\
& x=-2 \\
\end{aligned} \right.$
Do đó:
$S=\int\limits_{-2}^{2}{\left| {{x}^{3}}-4x \right|dx}=\int\limits_{-2}^{0}{\left| {{x}^{3}}-4x \right|dx}+\int\limits_{0}^{2}{\left| {{x}^{3}}-4x \right|dx}=\int\limits_{-2}^{0}{\left( {{x}^{3}}-4x \right)dx}+\int\limits_{0}^{2}{\left( {{x}^{3}}-4x \right)dx}$
$=\left( \dfrac{1}{4}{{x}^{4}}-2{{x}^{2}} \right)\left| _{\begin{smallmatrix}
\\
-2
\end{smallmatrix}}^{\begin{smallmatrix}
0 \\
\end{smallmatrix}} \right.-\left( \dfrac{1}{4}{{x}^{4}}-2{{x}^{2}} \right)\left| _{\begin{smallmatrix}
\\
0
\end{smallmatrix}}^{\begin{smallmatrix}
2 \\
\end{smallmatrix}} \right.=4+4=8.$
& x=0 \\
& x=2 \\
& x=-2 \\
\end{aligned} \right.$
Do đó:
$S=\int\limits_{-2}^{2}{\left| {{x}^{3}}-4x \right|dx}=\int\limits_{-2}^{0}{\left| {{x}^{3}}-4x \right|dx}+\int\limits_{0}^{2}{\left| {{x}^{3}}-4x \right|dx}=\int\limits_{-2}^{0}{\left( {{x}^{3}}-4x \right)dx}+\int\limits_{0}^{2}{\left( {{x}^{3}}-4x \right)dx}$
$=\left( \dfrac{1}{4}{{x}^{4}}-2{{x}^{2}} \right)\left| _{\begin{smallmatrix}
\\
-2
\end{smallmatrix}}^{\begin{smallmatrix}
0 \\
\end{smallmatrix}} \right.-\left( \dfrac{1}{4}{{x}^{4}}-2{{x}^{2}} \right)\left| _{\begin{smallmatrix}
\\
0
\end{smallmatrix}}^{\begin{smallmatrix}
2 \\
\end{smallmatrix}} \right.=4+4=8.$
| Công thức tính diện tích: $S=\int\limits_{{{x}_{0}}}^{{{x}_{n}}}{\left| f\left( x \right)-g\left( x \right) \right|dx}=\left| \int\limits_{{{x}_{0}}}^{{{x}_{1}}}{\left( f\left( x \right)-g\left( x \right) \right)dx} \right|+\left| \int\limits_{{{x}_{1}}}^{{{x}_{2}}}{\left( f\left( x \right)-g\left( x \right) \right)dx} \right|+...+\left| \int\limits_{{{x}_{n-1}}}^{{{x}_{n}}}{\left( f\left( x \right)-g\left( x \right) \right)dx} \right|$ |
Đáp án B.