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Câu hỏi: Cho hàm số $f\left( x \right)$ liên tục trên $\mathbb{R}$ sao cho $xf\left( {{x}^{3}} \right)+f\left( 1-{{x}^{2}} \right)=-{{x}^{8}}+2{{x}^{5}}-3x,\forall x\in \mathbb{R}$. Khi đó tích phân $\int_{-1}^{0}{f\left( x \right)dx}$ bằng
A. $\dfrac{579}{175}$.
B. $\dfrac{-17}{10}$.
C. $\dfrac{-13}{6}$.
D. $\dfrac{-579}{175}$.
A. $\dfrac{579}{175}$.
B. $\dfrac{-17}{10}$.
C. $\dfrac{-13}{6}$.
D. $\dfrac{-579}{175}$.
Ta có $xf\left( {{x}^{3}} \right)+f\left( 1-{{x}^{2}} \right)=-{{x}^{8}}+2{{x}^{5}}-3x$
$\Rightarrow {{x}^{2}}f\left( {{x}^{3}} \right)+xf\left( 1-{{x}^{2}} \right)=-{{x}^{9}}+2{{x}^{6}}-3{{x}^{2}}$
$\Rightarrow \int_{0}^{1}{{{x}^{2}}f\left( {{x}^{3}} \right)dx}+\int_{0}^{1}{xf\left( 1-{{x}^{2}} \right)dx}=\int_{0}^{1}{\left( -{{x}^{9}}+2{{x}^{6}}-3{{x}^{2}} \right)dx}$
$\Rightarrow \dfrac{1}{3}\int_{0}^{1}{f\left( t \right)dt}-\dfrac{1}{2}\int_{1}^{0}{f\left( u \right)du=\dfrac{-57}{70}}$ $\Rightarrow \dfrac{5}{6}\int_{0}^{1}{f\left( x \right)dx}=\dfrac{-57}{70}\Rightarrow \int_{0}^{1}{f\left( x \right)dx}=\dfrac{-171}{175}$
Ta lại có $\int_{-1}^{0}{{{x}^{2}}f\left( {{x}^{3}} \right)dx}+\int_{-1}^{0}{xf\left( 1-{{x}^{2}} \right)dx}=\int_{-1}^{0}{\left( -{{x}^{9}}+2{{x}^{6}}-3{{x}^{2}} \right)dx}$
$\Rightarrow \dfrac{1}{3}\int_{-1}^{0}{f\left( t \right)dt-}\dfrac{1}{2}\int_{1}^{0}{f\left( u \right)du=\dfrac{-43}{70}}$ $\Rightarrow \int_{-1}^{0}{f\left( x \right)dx}=3\left( \dfrac{-43}{70}+\dfrac{1}{2}.\dfrac{-171}{175} \right)=\dfrac{-579}{175}$
$\Rightarrow {{x}^{2}}f\left( {{x}^{3}} \right)+xf\left( 1-{{x}^{2}} \right)=-{{x}^{9}}+2{{x}^{6}}-3{{x}^{2}}$
$\Rightarrow \int_{0}^{1}{{{x}^{2}}f\left( {{x}^{3}} \right)dx}+\int_{0}^{1}{xf\left( 1-{{x}^{2}} \right)dx}=\int_{0}^{1}{\left( -{{x}^{9}}+2{{x}^{6}}-3{{x}^{2}} \right)dx}$
$\Rightarrow \dfrac{1}{3}\int_{0}^{1}{f\left( t \right)dt}-\dfrac{1}{2}\int_{1}^{0}{f\left( u \right)du=\dfrac{-57}{70}}$ $\Rightarrow \dfrac{5}{6}\int_{0}^{1}{f\left( x \right)dx}=\dfrac{-57}{70}\Rightarrow \int_{0}^{1}{f\left( x \right)dx}=\dfrac{-171}{175}$
Ta lại có $\int_{-1}^{0}{{{x}^{2}}f\left( {{x}^{3}} \right)dx}+\int_{-1}^{0}{xf\left( 1-{{x}^{2}} \right)dx}=\int_{-1}^{0}{\left( -{{x}^{9}}+2{{x}^{6}}-3{{x}^{2}} \right)dx}$
$\Rightarrow \dfrac{1}{3}\int_{-1}^{0}{f\left( t \right)dt-}\dfrac{1}{2}\int_{1}^{0}{f\left( u \right)du=\dfrac{-43}{70}}$ $\Rightarrow \int_{-1}^{0}{f\left( x \right)dx}=3\left( \dfrac{-43}{70}+\dfrac{1}{2}.\dfrac{-171}{175} \right)=\dfrac{-579}{175}$
Đáp án D.
