Tính tích phân 

D

- Hướng dẫn giải

Đáp án D.

Cách 1:

Ta có

$\begin{array}{l}{x}^3=x\\ \iff x\left(x-1\right)\left(x+1\right)=0\\ \iff \left[\begin{array}{l}x=0\\ x=\pm 1\end{array}\right.\end{array}$

 Do $x\in \left[0;2\right]$ nên $\left[\begin{array}{l}x=0\\ x=1\end{array}\right.$ 

Xét dấu, Ta được $x^3-x<\mathrm{0,}\forall x\in \left(0;1\right)$ và $x^3-x>\mathrm{0,}\forall x\in \left(1;2\right)$

Suy ra $\underset{\left[0;1\right]}{\max }\left\{x,x^3\right\}=x$ và $\underset{\left[1;2\right]}{\max }\left\{x,x^3\right\}=x^3.$

Vậy

$\begin{array}{l}\underset{0}{\overset{2}{\int }}\max \left\{x,x^3\right\}dx\\ =\underset{0}{\overset{1}{\int }}xdx+\underset{1}{\overset{2}{\int }}{x}^{3}dx=\frac{17}{4}\end{array}$ .

Cách 2: 

$\begin{array}{l}\underset{0}{\overset{2}{\int }}\max \left\{x,x^3\right\}dx\\ =\underset{0}{\overset{2}{\int }}\frac{x^3+x+\left|x^3-x\right|}{2}dx\\ =\underset{0}{\overset{1}{\int }}xdx+\underset{1}{\overset{2}{\int }}{x}^{3}dx=\frac{17}{4}.\end{array}$