Cho hàm số $f(x)$ xác định và liên tục trên $[0;+\infty)$ thỏa mãn $f(x)=x\sqrt{x}+\displaystyle\int\limits_{0}^{1}xf(x)\mathrm{\,d}x$. Tính tích phân $\displaystyle\int\limits_{0}^{4}f(x)\mathrm{\,d}x$.
 | $\dfrac{528}{35}$ |
 | $\dfrac{488}{35}$ |
 | $\dfrac{408}{35}$ |
 | $\dfrac{368}{35}$ |
Chọn phương án A.
Đặt $m=\displaystyle\int\limits_{0}^{1}xf(x)\mathrm{\,d}x$. Ta có $f(x)=x\sqrt{x}+m$. Khi đó $$\begin{aligned}
m&=\displaystyle\int\limits_{0}^{1}xf(x)\mathrm{\,d}x=\displaystyle\int\limits_{0}^{1}x\left(x\sqrt{x}+m\right)\mathrm{\,d}x\\
&=\displaystyle\int\limits_{0}^{1}\left(x^{\tfrac{5}{2}}+mx\right)\mathrm{\,d}x=\left(\dfrac{2}{7}x^{\tfrac{7}{2}}+\dfrac{mx^2}{2}\right)\bigg|_0^1\\
&=\dfrac{2}{7}+\dfrac{m}{2}.\\
\Leftrightarrow\dfrac{m}{2}&=\dfrac{2}{7}\Leftrightarrow m=\dfrac{4}{7}.
\end{aligned}$$
Vậy $f(x)=x\sqrt{x}+\dfrac{4}{7}$. Suy ra $$\displaystyle\int\limits_{0}^{4}f(x)\mathrm{\,d}x=\displaystyle\int\limits_{0}^{4}\left(x\sqrt{x}+\dfrac{4}{7}\right)\mathrm{\,d}x=\dfrac{528}{35}.$$