Cho hàm số $f(x)$ liên tục trên $\mathbb{R}$ thỏa mãn $f(x)=3x^2-2x+3+4\displaystyle\int\limits_{0}^{1}xf\left(x^2\right)\mathrm{\,d}x$. Khi đó $\displaystyle\int\limits_{2}^{3}f(x)\mathrm{\,d}x$ bằng
 | $17$ |
 | $11$ |
 | $14$ |
 | $21$ |
Chọn phương án B.
Đặt $m=\displaystyle\int\limits_{0}^{1}xf\left(x^2\right)\mathrm{\,d}x$.
Ta có $f(x)=3x^2-2x+3+4m$. Khi đó $$\begin{aligned}
m&=\displaystyle\int\limits_{0}^{1}xf\left(x^2\right)\mathrm{\,d}x=\displaystyle\int\limits_{0}^{1}x\left(3x^4-2x^2+3+4m\right)\mathrm{\,d}x\\
&=\displaystyle\int\limits_{0}^{1}\left(3x^5-2x^3+3x+4mx\right)\mathrm{\,d}x\\
&=\left(\dfrac{x^6}{2}-\dfrac{x^4}{2}+\dfrac{3x^2}{2}+2mx^2\right)\bigg|_0^1\\
&=\dfrac{1}{2}-\dfrac{1}{2}+\dfrac{3}{2}+2m=2m+\dfrac{3}{2}\\
\Leftrightarrow-m&=\dfrac{3}{2}\Leftrightarrow m=-\dfrac{3}{2}.
\end{aligned}$$
Vậy $f(x)=3x^2-2x-3$.
Suy ra $\displaystyle\int\limits_{2}^{3}\left(3x^2-2x-3\right)\mathrm{\,d}x=11$.