Tính tích phân sau 

D

- Hướng dẫn giải

Chọn D

Đặt $t=4-x^2\Rightarrow dt=-2xdx$$\Rightarrow xdx=-\frac{dt}{2}$

Khi x = 0 => t = 4; x = 2 => t = 0

$$\begin{gathered} D={\int }_4^0\left(-\frac{1}{2}\sqrt{t}\right)dt=\frac{1}{2}{\int }_0^4{t}^{\frac{1}{2}}dt \\ =\frac{1}{2}\left(\frac{2}{3}{t}^{\frac{3}{2}}\right)\overline{)\begin{array}{c}4\\ 0\end{array}=\frac{1}{3}\left(t\sqrt{t}\right)}\overline{)\begin{array}{c}4\\ 0\end{array}} \\ =\frac{1}{3}(4.2-0)=\frac{8}{3} \end{gathered}$$