Cho hàm số $f(x)$ xác định và liên tục trên đoạn $[0;1]$ thỏa mãn $f(x)=x^3+\displaystyle\int\limits_{0}^{1}x^3f\left(x^2\right)\mathrm{\,d}x$, $\forall x\in[0;1]$. Tính tích phân $\displaystyle\int\limits_{0}^{1}f(x)\mathrm{\,d}x$.
 | $\dfrac{1}{4}$ |
 | $\dfrac{4}{15}$ |
 | $\dfrac{13}{20}$ |
 | $\dfrac{23}{60}$ |
Chọn phương án D.
Đặt $m=\displaystyle\int\limits_{0}^{1}x^3f\left(x^2\right)\mathrm{\,d}x$. Ta có $f(x)=x^3+m$. Khi đó $$\begin{aligned}
m&=\displaystyle\int\limits_{0}^{1}x^3f\left(x^2\right)\mathrm{\,d}x=\displaystyle\int\limits_{0}^{1}x^3\left(x^6+m\right)\mathrm{\,d}x\\
&=\displaystyle\int\limits_{0}^{1}\left(x^9+mx^3\right)\mathrm{\,d}x=\left(\dfrac{x^{10}}{10}+\dfrac{mx^4}{4}\right)\bigg|_0^1\\
&=\dfrac{1}{10}+\dfrac{m}{4}\\
\Leftrightarrow\dfrac{3m}{4}&=\dfrac{1}{10}\Leftrightarrow m=\dfrac{2}{15}.
\end{aligned}$$
Vậy $f(x)=x^3+\dfrac{2}{15}$. Suy ra $$\displaystyle\int\limits_{0}^{1}f(x)\mathrm{\,d}x=\displaystyle\int\limits_{0}^{1}\left(x^3+\dfrac{2}{15}\right)\mathrm{\,d}x=\dfrac{23}{60}.$$