Ta có\[g\left( x \right) = 1 + 2\mathop \smallint \limits_0^x f\left( t \right)dt\]suy ra\(\left\{ {\begin{array}{*{20}{c}}{g(x) - 1 = 2\int\limits_0^2 {f(t)dt} }\\{g(0) = 1 + \int\limits_0^0 {f(t)dt} }\end{array}} \right.\)

\( \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{g\prime (x) = 2f(x) \Rightarrow f(x) = \frac{{g\prime (x)}}{2}}\\{g(0) = 1}\end{array}} \right.\)

Mà

\[g\left( x \right) \ge {\left[ {f\left( x \right)} \right]^3} \Leftrightarrow g\left( x \right) \ge {\left[ {\frac{{g'\left( x \right)}}{2}} \right]^3} \Leftrightarrow \sqrt[3]{{g\left( x \right)}} \ge \frac{{g'\left( x \right)}}{2} \Leftrightarrow \frac{{g'\left( x \right)}}{{\sqrt[3]{{g\left( x \right)}}}} \le 2\]

Với\[t \in \left[ {0;1} \right]\]Lấy tích phân hai vế ta được

\(\int\limits_0^t {\frac{{g\prime (x)}}{{\sqrt[3]{{g\left( x \right)}}}}} dx \le \int\limits_0^t {2dx} \)

\(\begin{array}{l} \Leftrightarrow \int\limits_0^t {{{[g(x)]}^{\frac{{ - 1}}{3}}}} d(g(x)) \le 2t\\ \Leftrightarrow 2t \ge 32{[g(x)]^{\frac{2}{3}}}\left| {_0^t} \right.\\ \Leftrightarrow \frac{4}{3}t \ge \sqrt[3]{{{g^2}(t)}} - \sqrt[3]{{{g^2}(0)}}\end{array}\)

Mà\[g\left( 0 \right) = 1\] nên\[\sqrt[3]{{{g^2}\left( t \right)}} \le \frac{4}{3}t + 1 \Rightarrow \sqrt[3]{{{g^2}\left( x \right)}} \le \frac{4}{3}x + 1\]

Từ đó ta có\[\mathop \smallint \limits_0^1 \sqrt[3]{{{g^2}\left( x \right)}}\,dx \le \mathop \smallint \limits_0^1 \left( {\frac{4}{3}x + 1} \right)dx\]

\( \Leftrightarrow \int\limits_0^1 {\sqrt[3]{{{g^2}(x)}}} dx \le \left( {\frac{2}{3}{x^2} + x} \right)\left| {_0^1} \right.\)

\( \Leftrightarrow \int\limits_0^1 {\sqrt[3]{{{g^2}(x)}}} dx \le \frac{5}{2}\)

Hay giá trị lớn nhất cần tìm là\[\frac{5}{3}.\]

Đáp án cần chọn là: B

Tích phân I = tích phân từ 1 đến 2 x 5 d x có giá trị là: