Cho hàm số y = f(x) có đạo hàm f'(x) liên tục trên đoạn [0; 1] thỏa mãn

Chọn B.

Đặt $\sqrt{x}=t\Rightarrow f\left(\sqrt{x}\right)=f\left(t\right)\Rightarrow f\text{'}\left(\sqrt{x}\right).\frac{1}{2\sqrt{x}}dx=f\text{'}\left(t\right)dt\Rightarrow f\text{'}\left(\sqrt{x}\right)dx=2tf\text{'}\left(t\right)dt$

Đổi cận: $x=0\Rightarrow t=0;x=1\Rightarrow t=1$

Khi đó: $\underset{0}{\overset{1}{\int }}f\text{'}\left(\sqrt{x}\right)dx=\underset{0}{\overset{1}{\int }}2tf\text{'}\left(t\right)dt=2\underset{0}{\overset{1}{\int }}tf\text{'}\left(t\right)dt$

Đặt $\left\{\begin{array}{l}u=t\\ dv=f\text{'}\left(t\right)dt\end{array}\right.\iff \left\{\begin{array}{l}du=dt\\ v=f\left(t\right)\end{array}\right.\Rightarrow 2\underset{0}{\overset{1}{\int }}tf\text{'}\left(t\right)dt=2tf\left(t\right)\left|\begin{array}{l}1\\ 0\end{array}\right.-2\underset{0}{\overset{1}{\int }}f\left(t\right)dt=2f\left(1\right)-4=-2.$