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Câu hỏi: Cho hàm số $y={{x}^{2}}-mx$ $\left( 0<m<2020 \right)$ có đồ thị $\left( C \right)$. Gọi ${{S}_{1}}+{{S}_{2}}$ là diện tích của hình phẳng giới hạn bởi $\left( C \right)$, trục hoành, trục tung và đường thẳng $x=2020$. Giá trị của $m$ sao cho ${{S}_{2}}={{S}_{1}}$ là
A. $m=\dfrac{4040}{3}$
B. $m=\dfrac{4041}{3}$
C. $m=\dfrac{2021}{3}$
D. $m=\dfrac{2020}{3}$
A. $m=\dfrac{4040}{3}$
B. $m=\dfrac{4041}{3}$
C. $m=\dfrac{2021}{3}$
D. $m=\dfrac{2020}{3}$
Ta có
${{S}_{2}}=\int\limits_{0}^{m}{\left( {{x}^{2}}-mx \right)dx}=\left. \left( \dfrac{{{x}^{3}}}{3}-\dfrac{m{{x}^{2}}}{2} \right) \right|_{m}^{2020}=\left( \dfrac{{{2020}^{3}}}{3}-\dfrac{m{{2020}^{2}}}{2} \right)+\dfrac{{{m}^{3}}}{6}$.
${{S}_{1}}=-\int\limits_{0}^{m}{\left( {{x}^{2}}-mx \right)dx}=-\left. \left( \dfrac{{{x}^{3}}}{3}-\dfrac{m{{x}^{2}}}{2} \right) \right|_{0}^{m}=\dfrac{{{m}^{3}}}{6}$.
${{S}_{2}}=2020{{S}_{1}}\Leftrightarrow \left( \dfrac{{{2020}^{3}}}{3}-\dfrac{m{{2020}^{2}}}{2} \right)+\dfrac{{{m}^{3}}}{6}=\dfrac{{{m}^{3}}}{6}$.
$\Leftrightarrow \dfrac{{{2020}^{3}}}{3}-\dfrac{m{{2020}^{2}}}{2}=0\Leftrightarrow m=\dfrac{4040}{3}$.
${{S}_{2}}=\int\limits_{0}^{m}{\left( {{x}^{2}}-mx \right)dx}=\left. \left( \dfrac{{{x}^{3}}}{3}-\dfrac{m{{x}^{2}}}{2} \right) \right|_{m}^{2020}=\left( \dfrac{{{2020}^{3}}}{3}-\dfrac{m{{2020}^{2}}}{2} \right)+\dfrac{{{m}^{3}}}{6}$.
${{S}_{1}}=-\int\limits_{0}^{m}{\left( {{x}^{2}}-mx \right)dx}=-\left. \left( \dfrac{{{x}^{3}}}{3}-\dfrac{m{{x}^{2}}}{2} \right) \right|_{0}^{m}=\dfrac{{{m}^{3}}}{6}$.
${{S}_{2}}=2020{{S}_{1}}\Leftrightarrow \left( \dfrac{{{2020}^{3}}}{3}-\dfrac{m{{2020}^{2}}}{2} \right)+\dfrac{{{m}^{3}}}{6}=\dfrac{{{m}^{3}}}{6}$.
$\Leftrightarrow \dfrac{{{2020}^{3}}}{3}-\dfrac{m{{2020}^{2}}}{2}=0\Leftrightarrow m=\dfrac{4040}{3}$.
Đáp án A.