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Câu hỏi: Cho $f\left( x \right)={{x}^{4}}-4{{x}^{3}}+2{{x}^{2}}-x+1.$ Tính $\int\limits_{0}^{1}{{{f}^{2}}\left( x \right){f}'\left( x \right)dx}?$
A. $-2.$
B. $\dfrac{2}{3}.$
C. 2.
D. $-\dfrac{2}{3}.$
A. $-2.$
B. $\dfrac{2}{3}.$
C. 2.
D. $-\dfrac{2}{3}.$
Ta có $\int\limits_{0}^{1}{{{f}^{2}}\left( x \right){f}'\left( x \right)dx}=\int\limits_{0}^{1}{{{f}^{2}}\left( x \right)d\left( f\left( x \right) \right)}=\dfrac{1}{3}\left. {{f}^{3}}\left( x \right) \right|_{0}^{1}=\dfrac{1}{3}\left[ {{\left( f\left( 1 \right) \right)}^{3}}-{{\left( f\left( 0 \right) \right)}^{3}} \right].$
Mà $f\left( 1 \right)=-1;f\left( 0 \right)=1.$ Do đó: $\int\limits_{0}^{1}{{{f}^{2}}\left( x \right){f}'\left( x \right)dx}=-\dfrac{2}{3}.$
Mà $f\left( 1 \right)=-1;f\left( 0 \right)=1.$ Do đó: $\int\limits_{0}^{1}{{{f}^{2}}\left( x \right){f}'\left( x \right)dx}=-\dfrac{2}{3}.$
Đáp án D.