Cho hàm số $f(x)$ có đạo hàm liên tục trên $\mathbb{R}$ thỏa mãn $f(x)=x^2-3x+2\displaystyle\int\limits_{0}^{1}f(x)f'(x)\mathrm{\,d}x$. Khi đó $\displaystyle\int\limits_{0}^{2}f(x)\mathrm{\,d}x$ bằng
 | $\dfrac{10}{3}$ |
 | $-\dfrac{10}{3}$ |
 | $\dfrac{26}{15}$ |
 | $-\dfrac{26}{15}$ |
Chọn phương án D.
Đặt $m=2\displaystyle\int\limits_{0}^{1}f(x)f'(x)\mathrm{\,d}x$.
Ta có $f(x)=x^2-3x+m$. Khi đó $$\begin{aligned}
m&=2\displaystyle\int\limits_{0}^{1}f(x)f'(x)\mathrm{\,d}x=\displaystyle\int\limits_{0}^{1}2f(x)f'(x)\mathrm{\,d}x\\ &=\displaystyle\int\limits_{0}^{1}\left[f^2(x)\right]'\mathrm{\,d}x=\left[f^2(x)\right]\bigg|_0^1\\ &=\left[x^2-3x+m\right]^2\bigg|_0^1\\
&=(-2+m)^2-m^2=4-4m\\
\Leftrightarrow5m&=4\Leftrightarrow m=\dfrac{4}{5}.
\end{aligned}$$
Vậy $f(x)=x^2-3x+\dfrac{4}{5}$. Suy ra $$\displaystyle\int\limits_{0}^{2}f(x)\mathrm{\,d}x=\displaystyle\int\limits_{0}^{2}\left(x^2-3x+\dfrac{4}{5}\right)\mathrm{\,d}x=-\dfrac{26}{15}.$$