Cho hàm số $f\left(x\right)$ thỏa mãn $f\left(2\right)=25$ và $f'\left(x\right)=4x\sqrt{f\left(x\right)}$ với mọi $x\in\mathbb{R}$. Khi đó $\displaystyle\displaystyle\int\limits_2^3f\left(x\right)\mathrm{\,d}x$ bằng
 | $\dfrac{1073}{15}$ |
 | $\dfrac{458}{15}$ |
 | $\dfrac{838}{15}$ |
 | $\dfrac{1016}{15}$ |
Chọn phương án C.
$\begin{aligned}
f'\left(x\right)=4x\sqrt{f\left(x\right)}\Rightarrow&\dfrac{f'\left(x\right)}{2\sqrt{f\left(x\right)}}=2x\\
\Rightarrow&\displaystyle\int\dfrac{f'\left(x\right)}{2\sqrt{f\left(x\right)}}\mathrm{\,d}x=\displaystyle\int2x\mathrm{\,d}x\\
\Rightarrow&\sqrt{f\left(x\right)}=x^2+C.
\end{aligned}$
Vì $f\left(2\right)=25$ nên $\sqrt{f\left(2\right)}=4+C\Rightarrow C=1$
Vậy $\sqrt{f\left(x\right)}=x^2+1\Rightarrow f\left(x\right)=\left(x^2+1\right)^2$
Khi đó $\displaystyle\int\limits_2^3f\left(x\right)\mathrm{\,d}x=\displaystyle\int\limits_2^3\left(x^2+1\right)^2\mathrm{\,d}x=\dfrac{838}{15}$.