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Câu hỏi: Cho hàm số $f\left( x \right)$ liên tục trên $\!\![\!\!0;1]$ thỏa mãn $f\left( x \right)=4{{x}^{3}}+k$ với $k=\int\limits_{0}^{1}{{{x}^{2}}f({{x}^{2}})}\text{d}x$. Khi đó $\int\limits_{0}^{1}{f(x)\text{d}x}$ bằng
A. $\dfrac{3}{2}.$
B. $\dfrac{5}{3}.$
C. $2.$
D. $\dfrac{2}{3}.$
A. $\dfrac{3}{2}.$
B. $\dfrac{5}{3}.$
C. $2.$
D. $\dfrac{2}{3}.$
Ta có: $k=\int\limits_{0}^{1}{{{x}^{2}}f({{x}^{2}})}\text{d}x=\int\limits_{0}^{1}{{{x}^{2}}(4{{x}^{6}}+k)}\text{d}x=\left. \left( \dfrac{4{{x}^{9}}}{9}+\dfrac{k{{x}^{3}}}{3} \right) \right|_{0}^{1}=\dfrac{4}{9}+\dfrac{k}{3}\Rightarrow k=\dfrac{2}{3}$
Do đó $\int\limits_{0}^{1}{f(x)\text{d}x}=\int\limits_{0}^{1}{\left( 4{{x}^{3}}+\dfrac{2}{3} \right)}\text{d}x=\dfrac{5}{3}.$
Do đó $\int\limits_{0}^{1}{f(x)\text{d}x}=\int\limits_{0}^{1}{\left( 4{{x}^{3}}+\dfrac{2}{3} \right)}\text{d}x=\dfrac{5}{3}.$
Đáp án B.
